RS Aggarwal Solutions for Class 12 Chapter 15 Integration Using Partial Fractions

Exercise 15A

Question 1. Evaluate: \( \int \frac{dx}{x(x+2)} \)
Answer: Let \( I = \int \frac{dx}{x(x+2)} \).

Set up the partial fractions as \( \frac{1}{x(x+2)} = \frac{A}{x} + \frac{B}{x+2} \).

This gives us \( A(x+2) + Bx = 1 \). When \( x = -2 \), we find \( B = -0.5 \). When \( x = 0 \), we get \( A = 0.5 \).

Therefore, \( \frac{1}{x(x+2)} = \frac{1}{2} \cdot \frac{1}{x} - \frac{1}{2} \cdot \frac{1}{x+2} \).

\( \int \frac{dx}{x(x+2)} = \frac{1}{2} \int \frac{dx}{x} - \frac{1}{2} \int \frac{dx}{x+2} = \frac{1}{2}\log|x| - \frac{1}{2}\log|x+2| + c \)

\( = \frac{1}{2}[\log|x| - \log|x+2|] + c = \frac{1}{2}\log\left|\frac{x}{x+2}\right| + c \)

Exam Tip: Always decompose the denominator into linear factors first, then apply the partial fraction method. Check your constants by substituting convenient values of x.

Question 2. Evaluate: \( \int \frac{(2x+1)dx}{(x+2)(x-3)} \)
Answer: Let \( I = \int \frac{(2x+1)dx}{(x+2)(x-3)} \).

Set up partial fractions as \( \frac{2x+1}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3} \). This gives us \( 2x+1 = A(x-3) + B(x+2) \).

When \( x = 3 \): \( 7 = 5B \), so \( B = \frac{7}{5} \).

When \( x = -2 \): \( -3 = -5A \), so \( A = \frac{3}{5} \).

Therefore, \( \frac{2x+1}{(x+2)(x-3)} = \frac{3}{5} \cdot \frac{1}{x+2} + \frac{7}{5} \cdot \frac{1}{x-3} \).

\( I = \frac{3}{5}\log|x+2| + \frac{7}{5}\log|x-3| + c \)

Exam Tip: The partial fraction setup must match the factored form of the denominator. Verify your values for A and B by comparing coefficients or substituting strategic x-values.

Question 3. Evaluate: \( \int \frac{x dx}{(x+2)(3-2x)} \)
Answer: Let \( I = \int \frac{x dx}{(x+2)(3-2x)} \).

Set up partial fractions as \( \frac{x}{(x+2)(3-2x)} = \frac{A}{x+2} + \frac{B}{3-2x} \). This gives us \( A(3-2x) + B(x+2) = x \).

When \( 3-2x = 0 \), we have \( x = \frac{3}{2} \). Then \( B\left(\frac{7}{2}\right) = \frac{3}{2} \), so \( B = \frac{3}{7} \).

When \( x = -2 \): \( A(7) = -2 \), so \( A = -\frac{2}{7} \).

Therefore, \( \frac{x}{(x+2)(3-2x)} = -\frac{2}{7} \cdot \frac{1}{x+2} + \frac{3}{7} \cdot \frac{1}{3-2x} \).

\( I = -\frac{2}{7}\log|x+2| + \frac{3}{7} \cdot \frac{1}{-2}\log|3-2x| + c = -\frac{2}{7}\log|x+2| - \frac{3}{14}\log|3-2x| + c \)

Exam Tip: When a linear term in the denominator has a coefficient other than 1, remember to account for it when integrating. The coefficient appears in the denominator of the final logarithm term.

Question 4. Evaluate: \( \int \frac{dx}{x(x-2)(x-4)} \)
Answer: Let \( I = \int \frac{dx}{x(x-2)(x-4)} \).

Set up partial fractions as \( \frac{1}{x(x-2)(x-4)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x-4} \). This gives us \( A(x-2)(x-4) + Bx(x-4) + Cx(x-2) = 1 \).

When \( x = 2 \): \( B(2)(-2) = 1 \), so \( B = -\frac{1}{4} \).

When \( x = 4 \): \( C(4)(2) = 1 \), so \( C = \frac{1}{8} \).

When \( x = 0 \): \( A(-2)(-4) = 1 \), so \( A = \frac{1}{8} \).

\( I = \frac{1}{8}\log|x| - \frac{1}{4}\log|x-2| + \frac{1}{8}\log|x-4| + c \)

Exam Tip: For three distinct linear factors in the denominator, use three separate constants. Substitute the roots to isolate each constant efficiently.

Question 5. Evaluate: \( \int \frac{(2x-1)dx}{(x-1)(x+2)(x-3)} \)
Answer: Let \( I = \int \frac{(2x-1)dx}{(x-1)(x+2)(x-3)} \).

Set up partial fractions as \( \frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3} \). This gives us \( A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2) = 2x-1 \).

When \( x = -2 \): \( B(-3)(-5) = -5 \), so \( B = -\frac{1}{3} \).

When \( x = 3 \): \( C(2)(5) = 5 \), so \( C = \frac{1}{2} \).

When \( x = 1 \): \( A(3)(-2) = 1 \), so \( A = -\frac{1}{6} \).

\( I = -\frac{1}{6}\log|x-1| - \frac{1}{3}\log|x+2| + \frac{1}{2}\log|x-3| + c \)

Exam Tip: Write the coefficients clearly when setting up the partial fraction equation. Substituting the roots eliminates all but one constant, making the calculation straightforward.

Question 6. Evaluate: \( \int \frac{(2x-3)dx}{(x^2-1)(2x+3)} \)
Answer: Let \( I = \int \frac{(2x-3)dx}{(x^2-1)(2x+3)} = \int \frac{(2x-3)dx}{(x-1)(x+1)(2x+3)} \).

Set up partial fractions as \( \frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3} \). This gives us \( A(x+1)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+1) = 2x-3 \).

When \( x = 1 \): \( A(2)(5) = -1 \), so \( A = -\frac{1}{10} \).

When \( x = -1 \): \( B(-2)(1) = -5 \), so \( B = \frac{5}{2} \).

When \( x = -\frac{3}{2} \): \( C\left(-\frac{5}{2}\right)\left(-\frac{1}{2}\right) = -6 \), so \( C = -\frac{24}{5} \).

\( I = -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{24}{5} \cdot \frac{1}{2}\log|2x+3| + c = -\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{12}{5}\log|2x+3| + c \)

Exam Tip: When a linear factor has a coefficient greater than 1, divide by that coefficient when integrating its reciprocal. Keep track of all terms to avoid sign errors.

Question 7. Evaluate: \( \int \frac{(2x+5)dx}{(x^2-x-2)} \)
Answer: Let \( I = \int \frac{(2x+5)dx}{(x^2-x-2)} = \int \frac{(2x+5)dx}{(x-2)(x+1)} \).

Set up partial fractions as \( \frac{2x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \). This gives us \( A(x+1) + B(x-2) = 2x+5 \).

When \( x = -1 \): \( B(-3) = 3 \), so \( B = -1 \).

When \( x = 2 \): \( A(3) = 9 \), so \( A = 3 \).

Therefore, \( \frac{2x+5}{(x-2)(x+1)} = \frac{3}{x-2} - \frac{1}{x+1} \).

\( I = 3\log|x-2| - \log|x+1| + c \)

Exam Tip: Always factor the denominator completely before setting up partial fractions. A quadratic can often be factored into linear terms, simplifying the decomposition.

Question 8. Evaluate: \( \int \frac{(x^2+5x+3)dx}{(x^2+3x+2)} \)
Answer: Let \( I = \int \frac{(x^2+5x+3)dx}{(x^2+3x+2)} \).

Since the degree of the numerator equals the degree of the denominator, perform polynomial long division. This yields \( \frac{x^2+5x+3}{x^2+3x+2} = 1 + \frac{2x+1}{x^2+3x+2} \).

Therefore, \( I = \int dx + \int \frac{2x+1}{(x+1)(x+2)} dx = x + I_1 \).

For \( I_1 \), set up partial fractions: \( \frac{2x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \). This gives us \( A(x+2) + B(x+1) = 2x+1 \).

When \( x = -2 \): \( B(-1) = -3 \), so \( B = 3 \).

When \( x = -1 \): \( A(1) = -1 \), so \( A = -1 \).

\( I_1 = -\log|x+1| + 3\log|x+2| + c \).

\( I = x - \log|x+1| + 3\log|x+2| + c \)

Exam Tip: When the numerator degree matches or exceeds the denominator degree, always divide first. The remainder will have a lower degree and can then be decomposed using partial fractions.

Question 9. Evaluate: \( \int \frac{(x^2+1)dx}{(x^2-1)} \)
Answer: Let \( I = \int \frac{(x^2+1)dx}{(x^2-1)} \).

Since the degree of the numerator equals the degree of the denominator, perform polynomial long division. This yields \( \frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1} \).

Therefore, \( I = \int dx + 2\int \frac{dx}{(x-1)(x+1)} \).

For the second integral, set up partial fractions: \( \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \). This gives us \( A(x+1) + B(x-1) = 1 \).

When \( x = 1 \): \( 2A = 1 \), so \( A = \frac{1}{2} \).

When \( x = -1 \): \( -2B = 1 \), so \( B = -\frac{1}{2} \).

\( I = x + 2 \left( \frac{1}{2}\log|x-1| - \frac{1}{2}\log|x+1| \right) + c = x + \log\left|\frac{x-1}{x+1}\right| + c \)

Exam Tip: Simplify first by dividing the polynomials, then decompose the remainder. This two-step approach avoids messy partial fraction work.

Question 10. Evaluate: \( \int \frac{x^5 dx}{(x^2-4)} \)
Answer: Let \( I = \int \frac{x^5 dx}{(x^2-4)} \).

Since the degree of the numerator exceeds the degree of the denominator, perform polynomial long division. This yields \( \frac{x^5}{x^2-4} = x^3 + 4x + \frac{4x}{x^2-4} \).

Therefore, \( I = \int x^3 dx + 4\int x dx + 4\int \frac{x dx}{x^2-4} \).

The first two integrals are standard: \( \int x^3 dx = \frac{x^4}{4} \) and \( 4\int x dx = 2x^2 \).

For the third integral, let \( u = x^2 - 4 \), so \( du = 2x dx \). Then \( \int \frac{4x dx}{x^2-4} = 2\int \frac{du}{u} = 2\log|u| = 2\log|x^2-4| \).

\( I = \frac{x^4}{4} + 2x^2 + 2\log|x^2-4| + c \)

Exam Tip: For high-degree numerators, polynomial division is essential. After division, you may have simpler integrals that don't require partial fractions at all.

Question 11. Evaluate: \( \int \frac{(5x+1)dx}{(x-1)(x+2)} \)
Answer: Let \( I = \int \frac{(5x+1)dx}{(x-1)(x+2)} \).

Set up partial fractions as \( \frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \). This gives us \( A(x+2) + B(x-1) = 5x+1 \).

When \( x = 1 \): \( A(3) = 6 \), so \( A = 2 \).

When \( x = -2 \): \( B(-3) = -9 \), so \( B = 3 \).

Therefore, \( \frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2} \).

\( I = 2\log|x-1| + 3\log|x+2| + c \)

Exam Tip: Verify your partial fraction decomposition by expanding the right side and checking that it matches the original numerator.

Question 12. Evaluate: \( \int \frac{x^3 dx}{(x-1)(x-2)} \)
Answer: Let \( I = \int \frac{x^3 dx}{(x-1)(x-2)} \).

Since the degree of the numerator exceeds the degree of the denominator, perform polynomial long division. This yields \( \frac{x^3}{(x-1)(x-2)} = x + 3 + \frac{7x-6}{(x-1)(x-2)} \).

Therefore, \( I = \int (x+3) dx + \int \frac{7x-6}{(x-1)(x-2)} dx = \frac{x^2}{2} + 3x + I_1 \).

For \( I_1 \), set up partial fractions: \( \frac{7x-6}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \). This gives us \( A(x-2) + B(x-1) = 7x-6 \).

When \( x = 2 \): \( B(1) = 8 \), so \( B = 8 \).

When \( x = 1 \): \( A(-1) = 1 \), so \( A = -1 \).

\( I_1 = -\log|x-1| + 8\log|x-2| + c \).

\( I = \frac{x^2}{2} + 3x - \log|x-1| + 8\log|x-2| + c \)

Exam Tip: Break the integration into manageable pieces. Polynomial long division simplifies the problem, and partial fractions handle the remainder systematically.

Question 13. Evaluate: \( \int \frac{(x^3-x-2)dx}{(1-x^2)} \)
Answer: Let \( I = \int \frac{(x^3-x-2)dx}{(1-x^2)} \).

Rewrite the denominator as \( 1 - x^2 = -(x^2-1) \), and rearrange: \( \frac{x^3-x-2}{1-x^2} = -x + \frac{-2}{1-x^2} \).

Therefore, \( I = \int (-x) dx + (-2) \int \frac{dx}{1-x^2} \).

The first integral is \( -\int x dx = -\frac{x^2}{2} \).

For the second, use the standard formula \( \int \frac{dx}{1-x^2} = \frac{1}{2}\log\left|\frac{1+x}{1-x}\right| + c \).

\( I = -\frac{x^2}{2} - 2 \cdot \frac{1}{2}\log\left|\frac{1+x}{1-x}\right| + c = -\frac{x^2}{2} + \log\left|\frac{1-x}{1+x}\right| + c \)

Exam Tip: Recognize standard integral forms like \( \int \frac{dx}{1-x^2} \). These appear frequently and save time if memorized.

Question 14. Evaluate: \( \int \frac{(2x+1)dx}{(4-3x-x^2)} \)
Answer: Let \( I = \int \frac{(2x+1)dx}{(4-3x-x^2)} = \int \frac{(2x+1)dx}{(1-x)(4+x)} \).

Set up partial fractions as \( \frac{2x+1}{(1-x)(4+x)} = \frac{A}{1-x} + \frac{B}{4+x} \). This gives us \( A(4+x) + B(1-x) = 2x+1 \).

When \( x = 1 \): \( A(5) = 3 \), so \( A = \frac{3}{5} \).

When \( x = -4 \): \( B(5) = -7 \), so \( B = -\frac{7}{5} \).

\( I = \frac{3}{5}\log|1-x| - \frac{7}{5}\log|4+x| + c = -\frac{1}{5}[3\log|1-x| + 7\log|4+x|] + c \)

Exam Tip: When the denominator is a quadratic, factor it into linear terms first. Be careful with the signs when the leading coefficient is negative.

Question 15. Evaluate: \( \int \frac{2x dx}{(x^2+1)(x^2-3)} \)
Answer: Let \( I = \int \frac{2x dx}{(x^2+1)(x^2-3)} \).

Use the substitution \( u = x^2 \), so \( du = 2x dx \). Then \( I = \int \frac{du}{(u+1)(u-3)} \).

Set up partial fractions as \( \frac{1}{(u+1)(u-3)} = \frac{A}{u+1} + \frac{B}{u-3} \). This gives us \( A(u-3) + B(u+1) = 1 \).

When \( u = 3 \): \( B(4) = 1 \), so \( B = \frac{1}{4} \).

When \( u = -1 \): \( A(-4) = 1 \), so \( A = -\frac{1}{4} \).

\( I = -\frac{1}{4}\log|u+1| + \frac{1}{4}\log|u-3| + c = \frac{1}{4}\log\left|\frac{u-3}{u+1}\right| + c = \frac{1}{4}\log\left|\frac{x^2-3}{x^2+1}\right| + c \)

Exam Tip: Substitution can convert a complex expression into a simpler form. Watch for patterns like \( 2x dx \) that correspond to derivatives of expressions involving \( x^2 \).

Question 16. Evaluate: \( \int \frac{\cos x dx}{(\cos^2 x - \cos x - 2)} \)
Answer: Let \( I = \int \frac{\cos x dx}{(\cos^2 x - \cos x - 2)} \).

Use the substitution \( t = \sin x \), so \( dt = \cos x dx \). Then \( \cos^2 x = 1 - t^2 \), giving us \( I = \int \frac{dt}{(1-t^2-t-2)} = \int \frac{dt}{(1+t)(2+t)} \).

Set up partial fractions as \( \frac{1}{(1+t)(2+t)} = \frac{A}{1+t} + \frac{B}{2+t} \). This gives us \( A(2+t) + B(1+t) = 1 \).

When \( t = -1 \): \( A(1) = 1 \), so \( A = 1 \).

When \( t = -2 \): \( B(-1) = 1 \), so \( B = -1 \).

\( I = \log|1+t| - \log|2+t| + c = \log\left|\frac{1+\sin x}{2+\sin x}\right| + c \)

Exam Tip: Trigonometric substitutions often simplify integrals involving trigonometric expressions. Choose the substitution that reduces the complexity most effectively.

Question 17. Evaluate: \( \int \frac{\sec^2 x dx}{(2+\tan x)(3+\tan x)} \)
Answer: Let \( I = \int \frac{\sec^2 x dx}{(2+\tan x)(3+\tan x)} \).

Use the substitution \( t = \tan x \), so \( dt = \sec^2 x dx \). Then \( I = \int \frac{dt}{(2+t)(3+t)} \).

Set up partial fractions as \( \frac{1}{(2+t)(3+t)} = \frac{A}{2+t} + \frac{B}{3+t} \). This gives us \( A(3+t) + B(2+t) = 1 \).

When \( t = -2 \): \( A(1) = 1 \), so \( A = 1 \).

When \( t = -3 \): \( B(-1) = 1 \), so \( B = -1 \).

\( I = \log|2+t| - \log|3+t| + c = \log\left|\frac{2+\tan x}{3+\tan x}\right| + c \)

Exam Tip: The pair \( \sec^2 x \) and \( \tan x \) is a natural match for substitution. Always look for these differential-derivative pairs.

Question 18. Evaluate: \( \int \frac{\sin x \cos x dx}{(\cos^2 x - \cos x - 2)} \)
Answer: Let \( I = \int \frac{\sin x \cos x dx}{(\cos^2 x - \cos x - 2)} \).

Use the substitution \( t = \cos x \), so \( dt = -\sin x dx \). Then \( I = \int \frac{-t dt}{(t^2 - t - 2)} = -\int \frac{t dt}{(t-2)(t+1)} \).

Set up partial fractions as \( \frac{t}{(t-2)(t+1)} = \frac{A}{t-2} + \frac{B}{t+1} \). This gives us \( A(t+1) + B(t-2) = t \).

When \( t = 2 \): \( A(3) = 2 \), so \( A = \frac{2}{3} \).

When \( t = -1 \): \( B(-3) = -1 \), so \( B = \frac{1}{3} \).

\( I = -\left( \frac{2}{3}\log|t-2| + \frac{1}{3}\log|t+1| \right) + c = -\frac{2}{3}\log|\cos x - 2| - \frac{1}{3}\log|\cos x + 1| + c \)

Exam Tip: When a trigonometric integral contains a product like \( \sin x \cos x \), use substitution with one of the trigonometric terms to simplify.

Question 19. Evaluate: \( \int \frac{e^x dx}{(e^{2x}+5e^x+6)} \)
Answer: Let \( I = \int \frac{e^x dx}{(e^{2x}+5e^x+6)} \).

Use the substitution \( t = e^x \), so \( dt = e^x dx \). Then \( I = \int \frac{dt}{(t^2+5t+6)} = \int \frac{dt}{(t+2)(t+3)} \).

Set up partial fractions as \( \frac{1}{(t+2)(t+3)} = \frac{A}{t+2} + \frac{B}{t+3} \). This gives us \( A(t+3) + B(t+2) = 1 \).

When \( t = -2 \): \( A(1) = 1 \), so \( A = 1 \).

When \( t = -3 \): \( B(-1) = 1 \), so \( B = -1 \).

\( I = \log|t+2| - \log|t+3| + c = \log\left|\frac{e^x+2}{e^x+3}\right| + c \)

Exam Tip: Exponential integrals often simplify with substitution. The substitution \( t = e^x \) converts the integral into a rational function that can be handled with partial fractions.

Question 20. Evaluate: \( \int \frac{e^x dx}{(e^{2x}-3e^x-e^x+3)} \)
Answer: Let \( I = \int \frac{e^x dx}{(e^{2x}-3e^x-e^x+3)} = \int \frac{e^x dx}{(e^{2x}-4e^x+3)} \).

Use the substitution \( t = e^x \), so \( dt = e^x dx \). Then \( I = \int \frac{dt}{(t^2-4t+3)} = \int \frac{dt}{(t-1)(t-3)} \).

Set up partial fractions as \( \frac{1}{(t-1)(t-3)} = \frac{A}{t-1} + \frac{B}{t-3} \). This gives us \( A(t-3) + B(t-1) = 1 \).

When \( t = 1 \): \( A(-2) = 1 \), so \( A = -\frac{1}{2} \).

When \( t = 3 \): \( B(2) = 1 \), so \( B = \frac{1}{2} \).

\( I = -\frac{1}{2}\log|t-1| + \frac{1}{2}\log|t-3| + c = \frac{1}{2}\log\left|\frac{e^x-3}{e^x-1}\right| + c = \log\left|\frac{e^x+2}{e^x+3}\right| + c \)

Exam Tip: Always factor the denominator completely. A quadratic in \( e^x \) often factors into linear terms, making the partial fraction decomposition straightforward.

Question 21. Evaluate: \( \int \frac{2\log x}{x[2(\log x)^2 - \log x - 3]} dx \)
Answer: Let \( I = \int \frac{2\log x}{x[2(\log x)^2 - \log x - 3]} dx \).

Put \( t = \log x \), so \( dt = \frac{dx}{x} \).

Then \( I = \int \frac{2t}{2t^2 - t - 3} dt \).

Using partial fractions, decompose \( \frac{2t}{2t^2 - t - 3} = \frac{A}{2t - 3} + \frac{B}{t + 1} \).

This gives \( A(t + 1) + B(2t - 3) = 2t \).

Setting \( 2t - 3 = 0 \) yields \( t = \frac{3}{2} \), so \( A\left(\frac{5}{2}\right) = 3 \), giving \( A = \frac{6}{5} \).

Setting \( t + 1 = 0 \) yields \( t = -1 \), so \( B(-5) = -2 \), giving \( B = \frac{2}{5} \).

Therefore, \( \frac{2t}{2t^2 - t - 3} = \frac{6}{5} \cdot \frac{1}{2t - 3} + \frac{2}{5} \cdot \frac{1}{t + 1} \).

\( I = \frac{6}{5} \int \frac{dt}{2t - 3} + \frac{2}{5} \int \frac{dt}{t + 1} = \frac{6}{5} \cdot \frac{\log|2t - 3|}{2} + \frac{2}{5} \log|t + 1| + c \)

\( = \frac{3}{5} \log|2\log x - 3| + \frac{2}{5} \log|\log x + 1| + c \)

In simple words: Use substitution to change the variable to \( t = \log x \), then apply partial fraction decomposition to split the rational function into simpler pieces. Integrate each piece separately and substitute back the original variable.

Exam Tip: Always check partial fraction coefficients by substituting specific values - this catches arithmetic errors quickly before integration.

Question 22. Evaluate: \( \int \frac{\operatorname{cosec}^2 x}{1 - \cot^2 x} dx \)
Answer: Let \( I = \int \frac{\operatorname{cosec}^2 x}{1 - \cot^2 x} dx \).

Put \( t = \cot x \), so \( dt = -\operatorname{cosec}^2 x \, dx \).

Then \( I = \int \frac{-dt}{1 - t^2} = -\int \frac{dt}{1 - t^2} \).

Using partial fractions, \( \frac{1}{1 - t^2} = \frac{1}{(1-t)(1+t)} = \frac{1/2}{1 - t} + \frac{1/2}{1 + t} \).

\( I = -\frac{1}{2} \int \frac{dt}{1 - t} - \frac{1}{2} \int \frac{dt}{1 + t} = \frac{1}{2} \log|1 - t| - \frac{1}{2} \log|1 + t| + c \)

\( = \frac{1}{2} \log\left|\frac{1 - \cot x}{1 + \cot x}\right| + c \)

In simple words: Replace \( \cot x \) with a new variable, then use partial fractions to break down the result into logarithmic integrals.

Exam Tip: Remember that \( \operatorname{cosec}^2 x \) is the derivative of \( -\cot x \), making the substitution natural and reducing computational effort.

Question 23. Evaluate: \( \int \frac{\sec^2 x}{(\tan^2 x + 4\tan x)} dx \)
Answer: Let \( I = \int \frac{\sec^2 x}{(\tan^2 x + 4\tan x)} dx \).

Put \( t = \tan x \), so \( dt = \sec^2 x \, dx \).

Then \( I = \int \frac{dt}{t^2 + 4t} = \int \frac{dt}{t(t + 4)} \).

Using partial fractions, decompose \( \frac{1}{t(t + 4)} = \frac{A}{t} + \frac{B}{t + 4} \).

This gives \( A(t + 4) + Bt = 1 \).

Setting \( t = 0 \) yields \( 4A = 1 \), so \( A = \frac{1}{4} \).

Setting \( t = -4 \) yields \( -4B = 1 \), so \( B = -\frac{1}{4} \).

Therefore, \( \frac{1}{t(t + 4)} = \frac{1}{4} \cdot \frac{1}{t} - \frac{1}{4} \cdot \frac{1}{t + 4} \).

\( I = \frac{1}{4} \log|t| - \frac{1}{4} \log|t + 4| + c = \frac{1}{4} \log\left|\frac{t}{t + 4}\right| + c = \frac{1}{4} \log\left|\frac{\tan x}{\tan x + 4}\right| + c \)

In simple words: Substitute \( t = \tan x \) to convert the trigonometric integral into a rational function, then decompose using partial fractions and integrate to get logarithmic terms.

Exam Tip: Always verify that the denominator factors completely before setting up partial fractions - this ensures your decomposition form is correct.

Question 24. Evaluate: \( \int \frac{\sin 2x}{(1 + \sin x)(2 + \sin x)} dx \)
Answer: Let \( I = \int \frac{\sin 2x}{(1 + \sin x)(2 + \sin x)} dx \).

Put \( t = \sin x \), so \( dt = \cos x \, dx \). Since \( \sin 2x = 2\sin x \cos x = 2t\sqrt{1 - t^2} \) (or equivalently, \( dt = \cos x \, dx \) and we rewrite using \( \sin 2x = 2\sin x \cos x \)), we have:

\( I = \int \frac{2t}{(1 + t)(2 + t)} dt \).

Using partial fractions, decompose \( \frac{2t}{(1 + t)(2 + t)} = \frac{A}{1 + t} + \frac{B}{2 + t} \).

This gives \( A(2 + t) + B(1 + t) = 2t \).

Setting \( t = -1 \) yields \( A(1) = -2 \), so \( A = -2 \).

Setting \( t = -2 \) yields \( B(-1) = -4 \), so \( B = 4 \).

Therefore, \( \frac{2t}{(1 + t)(2 + t)} = -\frac{2}{1 + t} + \frac{4}{2 + t} \).

\( I = -2\int \frac{dt}{1 + t} + 4\int \frac{dt}{2 + t} = -2\log|1 + t| + 4\log|2 + t| + c \)

\( = 4\log|2 + \sin x| - 2\log|1 + \sin x| + c \)

In simple words: Set \( t = \sin x \) to change the integral into a rational function, then split it using partial fractions. Integrate the resulting logarithmic terms and substitute back.

Exam Tip: When the coefficient pattern suggests particular roots (like 1 and -2 here), set up partial fractions immediately and use root substitution to find coefficients efficiently.

Question 25. Evaluate: \( \int \frac{e^x}{e^x(e^x - 1)} dx \)
Answer: Let \( I = \int \frac{e^x}{e^x(e^x - 1)} dx = \int \frac{1}{e^x - 1} dx \).

Put \( t = e^x \), so \( dt = e^x \, dx \), giving \( dx = \frac{dt}{t} \).

Then \( I = \int \frac{1}{t(t - 1)} \cdot \frac{dt}{t} = \int \frac{dt}{t(t - 1)} \).

Using partial fractions, decompose \( \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \).

This gives \( A(t - 1) + Bt = 1 \).

Setting \( t = 0 \) yields \( -A = 1 \), so \( A = -1 \).

Setting \( t = 1 \) yields \( B = 1 \).

Therefore, \( \frac{1}{t(t - 1)} = -\frac{1}{t} + \frac{1}{t - 1} \).

\( I = -\int \frac{dt}{t} + \int \frac{dt}{t - 1} = -\log|t| + \log|t - 1| + c = \log\left|\frac{t - 1}{t}\right| + c \)

\( = \log\left|\frac{e^x - 1}{e^x}\right| + c \)

In simple words: Replace \( e^x \) with a new variable to convert the exponential integral into a simpler rational form. Use partial fractions to separate it into basic logarithmic integrals.

Exam Tip: Always simplify the integrand first before substitution - here the \( e^x \) cancels immediately, avoiding unnecessary complexity.

Question 26. Evaluate: \( \int \frac{dx}{x(x^4 - 1)} \)
Answer: Let \( I = \int \frac{dx}{x(x^4 - 1)} dx \).

Put \( t = x^4 \), so \( dt = 4x^3 \, dx \), giving \( dx = \frac{dt}{4x^3} = \frac{x^3 \, dt}{4x^4} = \frac{dt}{4x^3} \).

Note that \( x^4 \, dx = \frac{1}{4} \cdot dt \cdot x \), but it's cleaner to write:

\( I = \int \frac{dx}{x(x^4 - 1)} = \int \frac{x^3 \, dx}{x^4(x^4 - 1)} = \frac{1}{4} \int \frac{dt}{t(t - 1)} \).

Using partial fractions, \( \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \).

This gives \( A(t - 1) + Bt = 1 \).

Setting \( t = 0 \) yields \( A = -1 \).

Setting \( t = 1 \) yields \( B = 1 \).

Therefore, \( \frac{1}{t(t - 1)} = -\frac{1}{t} + \frac{1}{t - 1} \).

\( I = \frac{1}{4}\left(-\int \frac{dt}{t} + \int \frac{dt}{t - 1}\right) = \frac{1}{4}(-\log|t| + \log|t - 1|) + c = \frac{1}{4}\log\left|\frac{t - 1}{t}\right| + c \)

\( = \frac{1}{4}\log\left|\frac{x^4 - 1}{x^4}\right| + c = -\log|x| + \frac{1}{4}\log|x^4 - 1| + c \)

In simple words: Use substitution to replace the fourth power expression with a single variable, convert to partial fractions, and integrate. Then back-substitute to return to the original variable.

Exam Tip: Always scale the logarithmic result correctly - the constant \( \frac{1}{4} \) multiplying the integral will appear in the final answer.

Question 41. Evaluate: \( \int \frac{dx}{x^3 - 1} \)
Answer: Let \( I = \int \frac{dx}{x^3 - 1} dx \).

Factor the denominator: \( x^3 - 1 = (x - 1)(x^2 + x + 1) \).

Using partial fractions, decompose \( \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \).

This gives \( A(x^2 + x + 1) + (Bx + C)(x - 1) = 1 \).

Setting \( x = 1 \) yields \( 3A = 1 \), so \( A = \frac{1}{3} \).

By comparing coefficients of \( x^2 \): \( A + B = 0 \), giving \( B = -\frac{1}{3} \).

By comparing constant terms: \( A - C = 1 \), giving \( C = A - 1 = -\frac{2}{3} \).

Therefore, \( \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{1}{3} \cdot \frac{1}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1} \).

\( I = \frac{1}{3}\int \frac{dx}{x - 1} - \frac{1}{3}\int \frac{x}{x^2 + x + 1} dx - \frac{2}{3}\int \frac{dx}{x^2 + x + 1} \).

For the second integral, use \( t = x^2 + x + 1 \), so \( dt = (2x + 1)dx \):

\( \int \frac{x}{x^2 + x + 1} dx = \frac{1}{2}\int \frac{2x + 1 - 1}{x^2 + x + 1} dx = \frac{1}{2}\log|x^2 + x + 1| - \frac{1}{2}\int \frac{dx}{x^2 + x + 1} \).

For the integral \( \int \frac{dx}{x^2 + x + 1} \), complete the square: \( x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \).

\( \int \frac{dx}{x^2 + x + 1} = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) \).

Combining all parts:

\( I = \frac{1}{3}\log|x - 1| - \frac{1}{6}\log|x^2 + x + 1| + \frac{1}{6}\int \frac{dx}{x^2 + x + 1} - \frac{2}{3}\int \frac{dx}{x^2 + x + 1} \)

\( = \frac{1}{3}\log|x - 1| - \frac{1}{6}\log|x^2 + x + 1| - \frac{1}{2}\int \frac{dx}{x^2 + x + 1} \)

\( = \frac{1}{3}\log|x - 1| - \frac{1}{6}\log|x^2 + x + 1| - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + c \)

In simple words: Factor the cubic denominator, set up partial fractions with a linear term and a quadratic term, integrate each piece separately using substitution and completing the square where needed.

Exam Tip: When integrating over an irreducible quadratic, always complete the square first - this reveals the arctangent form immediately.

Question 42. Evaluate: \( \int \frac{dx}{x^3 + 1} \)
Answer: Let \( I = \int \frac{dx}{x^3 + 1} dx \).

Factor the denominator: \( x^3 + 1 = (x + 1)(x^2 - x + 1) \).

Using partial fractions, decompose \( \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} \).

This gives \( A(x^2 - x + 1) + (Bx + C)(x + 1) = 1 \).

Setting \( x = -1 \) yields \( A(1 + 1 + 1) = 1 \), so \( A = \frac{1}{3} \).

By comparing coefficients of \( x^2 \): \( A + B = 0 \), giving \( B = -\frac{1}{3} \).

By comparing constant terms: \( A + C = 1 \), giving \( C = \frac{2}{3} \).

Therefore, \( \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{1}{3} \cdot \frac{1}{x + 1} + \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} \).

\( I = \frac{1}{3}\log|x + 1| - \frac{1}{3}\int \frac{x - 2}{x^2 - x + 1} dx \).

For the integral \( \int \frac{x - 2}{x^2 - x + 1} dx \), rewrite the numerator: \( x - 2 = \frac{1}{2}(2x - 1) - \frac{3}{2} \).

\( \int \frac{x - 2}{x^2 - x + 1} dx = \frac{1}{2}\int \frac{2x - 1}{x^2 - x + 1} dx - \frac{3}{2}\int \frac{dx}{x^2 - x + 1} \)

\( = \frac{1}{2}\log|x^2 - x + 1| - \frac{3}{2}\int \frac{dx}{x^2 - x + 1} \).

For \( \int \frac{dx}{x^2 - x + 1} \), complete the square: \( x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} \).

\( \int \frac{dx}{x^2 - x + 1} = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) \).

Therefore:

\( I = \frac{1}{3}\log|x + 1| - \frac{1}{6}\log|x^2 - x + 1| + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{3}}\right) + c \)

In simple words: Decompose the cubic into linear and quadratic factors, apply partial fractions with appropriate numerators, and integrate each term. Complete the square on the irreducible quadratic to find the arctangent integral.

Exam Tip: When the numerator of a quadratic fraction is linear, separate it into a derivative term (for logarithm) and a constant term (for arctangent) using inspection.

Question 27. Evaluate: \( \int \frac{(1 - x^2)}{x(1 - 2x)} dx \)
Answer: Let \( I = \int \frac{1 - x^2}{x(1 - 2x)} dx \).

Rewrite the numerator: \( 1 - x^2 = \frac{1}{2} + \frac{x - 2x^2}{2(1 - 2x)} \) (by polynomial division or clever rearrangement).

Actually, let's decompose directly. We have:

\( \int \frac{1 - x^2}{x(1 - 2x)} dx = \int \left(\frac{1}{2} + \frac{x}{x(1 - 2x)} - \frac{1}{x(1 - 2x)}\right) dx \)

Let \( I_1 = \int \frac{1}{x(1 - 2x)} dx \). Using partial fractions:

\( \frac{1}{x(1 - 2x)} = \frac{A}{x} + \frac{B}{1 - 2x} \).

This gives \( A(1 - 2x) + Bx = 1 \).

Setting \( x = 0 \) yields \( A = 1 \).

Setting \( 1 - 2x = 0 \) (i.e., \( x = \frac{1}{2} \)) yields \( B\left(\frac{1}{2}\right) = 1 \), so \( B = 2 \).

Therefore, \( I_1 = \int \frac{dx}{x} + 2\int \frac{dx}{1 - 2x} = \log|x| - \log|1 - 2x| \).

Combining:

\( I = \frac{1}{2}x + \frac{1}{4}\log|2x - 1| - \log|2x - 1| + \log|x| + c \)

\( = \frac{1}{2}x - \frac{3}{4}\log|1 - 2x| + \log|x| + c \)

In simple words: Split the rational integrand using polynomial division and partial fractions, then integrate each simpler piece separately using standard logarithmic formulas.

Exam Tip: Partial fractions work best when the degree of the numerator is less than the denominator - if not, perform polynomial long division first.

Question 45. Evaluate: \( \int \frac{dx}{(x^2 + 2)(x^2 + 4)} \)
Answer: Let \( I = \int \frac{dx}{(x^2 + 2)(x^2 + 4)} dx \).

Using partial fractions, decompose \( \frac{1}{(x^2 + 2)(x^2 + 4)} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 4} \).

This gives \( A(x^2 + 4) + B(x^2 + 2) = 1 \).

Setting \( x^2 = -2 \) yields \( A(2) = 1 \), so \( A = \frac{1}{2} \).

Setting \( x^2 = -4 \) yields \( B(-2) = 1 \), so \( B = -\frac{1}{2} \).

Therefore, \( \frac{1}{(x^2 + 2)(x^2 + 4)} = \frac{1}{2} \cdot \frac{1}{x^2 + 2} - \frac{1}{2} \cdot \frac{1}{x^2 + 4} \).

\( I = \frac{1}{2}\int \frac{dx}{x^2 + 2} - \frac{1}{2}\int \frac{dx}{x^2 + 4} \).

Using the formula \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \):

\( I = \frac{1}{2} \cdot \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) - \frac{1}{2} \cdot \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \)

\( = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) - \frac{1}{4}\tan^{-1}\left(\frac{x}{2}\right) + c \)

In simple words: Break the product of two quadratics into a sum using partial fractions, then apply the standard arctangent integral formula to each term.

Exam Tip: For sums of squared terms in the denominator, partial fractions always work even though there are no real roots - use the complex root method or substitution method.

Question 28. Evaluate: \( \int \frac{(x^2 + x + 1)}{(x + 2)(x + 1)^2} dx \)
Answer: Let \( I = \int \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} dx \).

Using partial fractions, decompose \( \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} = \frac{A}{x + 2} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \).

This gives \( A(x + 1)^2 + B(x + 2)(x + 1) + C(x + 2) = x^2 + x + 1 \).

Setting \( x = -1 \) yields \( C(1) = 1 - 1 + 1 = 1 \), so \( C = 1 \).

Setting \( x = -2 \) yields \( A(1) = 4 - 2 + 1 = 3 \), so \( A = 3 \).

Comparing coefficients of \( x^2 \): \( A + B = 1 \), so \( B = -2 \).

Therefore, \( \frac{x^2 + x + 1}{(x + 2)(x + 1)^2} = \frac{3}{x + 2} - \frac{2}{x + 1} + \frac{1}{(x + 1)^2} \).

\( I = 3\int \frac{dx}{x + 2} - 2\int \frac{dx}{x + 1} + \int \frac{dx}{(x + 1)^2} \)

\( = 3\log|x + 2| - 2\log|x + 1| - \frac{1}{x + 1} + c \)

In simple words: When a factor in the denominator appears with a power greater than 1, include all powers from 1 up to that maximum in the partial fraction decomposition. Integrate each term using standard formulas.

Exam Tip: Always check repeated linear factors carefully - each power must have its own term in the decomposition.

Question 43. Evaluate: \( \int \frac{dx}{(x^2 + 1)(x + 1)^2} \)
Answer: Let \( I = \int \frac{dx}{(x^2 + 1)(x + 1)^2} dx \).

Using partial fractions, decompose \( \frac{1}{(x^2 + 1)(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{Cx + D}{x^2 + 1} \).

This gives \( A(x + 1)(x^2 + 1) + B(x^2 + 1) + (Cx + D)(x + 1)^2 = 1 \).

Setting \( x = -1 \) yields \( B(2) = 1 \), so \( B = \frac{1}{2} \).

Comparing coefficients of \( x^2 \): \( A + C = 0 \), so \( C = -A \).

Comparing coefficients of \( x^3 \): \( A + 2C = 0 \), which with \( C = -A \) gives \( A - 2A = 0 \), so \( A = 0 \) and \( C = 0 \).

Comparing constant terms: \( A + B + D = 1 \), so \( \frac{1}{2} + D = 1 \), giving \( D = \frac{1}{2} \).

Therefore, \( \frac{1}{(x^2 + 1)(x + 1)^2} = \frac{1}{2} \cdot \frac{1}{(x + 1)^2} + \frac{1}{2} \cdot \frac{1}{x^2 + 1} \).

\( I = \frac{1}{2}\int \frac{dx}{(x + 1)^2} + \frac{1}{2}\int \frac{dx}{x^2 + 1} \)

\( = -\frac{1}{2} \cdot \frac{1}{x + 1} + \frac{1}{2}\tan^{-1}(x) + c \)

In simple words: Set up partial fractions carefully with a repeated linear factor (two terms) and an irreducible quadratic (linear numerator). Integrate each piece using basic formulas.

Exam Tip: When comparing coefficients, organize by powers systematically - this reduces errors and reveals which variables can be eliminated immediately.

Question 44. Evaluate: \( \int \frac{17}{(2x + 1)(x^2 + 4)} dx \)
Answer: Let \( I = \int \frac{17}{(2x + 1)(x^2 + 4)} dx \).

Using partial fractions, decompose \( \frac{17}{(2x + 1)(x^2 + 4)} = \frac{A}{2x + 1} + \frac{Bx + C}{x^2 + 4} \).

This gives \( A(x^2 + 4) + (Bx + C)(2x + 1) = 17 \).

Setting \( 2x + 1 = 0 \) (i.e., \( x = -\frac{1}{2} \)) yields \( A\left(\frac{1}{4} + 4\right) = 17 \), so \( A\left(\frac{17}{4}\right) = 17 \), giving \( A = 4 \).

Comparing coefficients of \( x^2 \): \( A + 2B = 0 \), so \( B = -2 \).

Comparing constant terms: \( 4A + C = 17 \), so \( C = 1 \).

Therefore, \( \frac{17}{(2x + 1)(x^2 + 4)} = \frac{4}{2x + 1} + \frac{-2x + 1}{x^2 + 4} \).

\( I = 4\int \frac{dx}{2x + 1} - 2\int \frac{x \, dx}{x^2 + 4} + \int \frac{dx}{x^2 + 4} \)

\( = 2\log|2x + 1| - \log|x^2 + 4| + \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c \)

In simple words: Decompose the rational function with a linear and a quadratic factor by setting up partial fractions with appropriate numerators. Integrate each piece using logarithmic and arctangent formulas.

Exam Tip: When integrating \( \frac{x}{x^2 + a^2} \), recognize it as half the derivative of the denominator - this yields a logarithm directly.

Question 29. Evaluate: \( \int \frac{(2x + 9)}{(x + 2)(x - 3)^2} dx \)
Answer: Let \( I = \int \frac{2x + 9}{(x + 2)(x - 3)^2} dx \).

Using partial fractions, decompose \( \frac{2x + 9}{(x + 2)(x - 3)^2} = \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2} \).

This gives \( A(x - 3)^2 + B(x + 2)(x - 3) + C(x + 2) = 2x + 9 \).

Setting \( x = 3 \) yields \( C(5) = 15 \), so \( C = 3 \).

Setting \( x = -2 \) yields \( A(25) = 5 \), so \( A = \frac{1}{5} \).

Comparing coefficients of \( x^2 \): \( A + B = 0 \), so \( B = -\frac{1}{5} \).

Therefore, \( \frac{2x + 9}{(x + 2)(x - 3)^2} = \frac{1}{5} \cdot \frac{1}{x + 2} - \frac{1}{5} \cdot \frac{1}{x - 3} + \frac{3}{(x - 3)^2} \).

\( I = \frac{1}{5}\int \frac{dx}{x + 2} - \frac{1}{5}\int \frac{dx}{x - 3} + 3\int \frac{dx}{(x - 3)^2} \)

\( = \frac{1}{5}\log|x + 2| - \frac{1}{5}\log|x - 3| - \frac{3}{x - 3} + c \)

In simple words: Include a term for each power of the repeated factor in the decomposition. Use substitution values to find the coefficients, then integrate each term separately using standard formulas.

Exam Tip: For a repeated factor like \( (x - a)^n \), always include all powers: \( \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \ldots + \frac{A_n}{(x - a)^n} \).

Question 46. Evaluate: \( \int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} dx \)
Answer: Let \( I = \int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} dx \).

Using partial fractions, decompose \( \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} = \frac{A}{x^2 + 4} + \frac{B}{x^2 + 25} \).

This gives \( A(x^2 + 25) + B(x^2 + 4) = x^2 + 1 \).

Setting \( x^2 = -4 \) yields \( A(21) = -3 \), so \( A = -\frac{1}{7} \).

Setting \( x^2 = -25 \) yields \( B(-21) = -24 \), so \( B = \frac{8}{7} \).

Therefore, \( \frac{x^2 + 1}{(x^2 + 4)(x^2 + 25)} = -\frac{1}{7} \cdot \frac{1}{x^2 + 4} + \frac{8}{7} \cdot \frac{1}{x^2 + 25} \).

\( I = -\frac{1}{7}\int \frac{dx}{x^2 + 4} + \frac{8}{7}\int \frac{dx}{x^2 + 25} \)

\( = -\frac{1}{7} \cdot \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + \frac{8}{7} \cdot \frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right) + c \)

\( = -\frac{1}{14}\tan^{-1}\left(\frac{x}{2}\right) + \frac{8}{35}\tan^{-1}\left(\frac{x}{5}\right) + c \)

In simple words: Use partial fractions to decompose a ratio of quadratics, apply the arctangent integral formula to each piece with the correct scaling factor.

Exam Tip: For quadratic factors with no real roots, the formula \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \) works directly - just match the denominator to this form.

Question 47. Evaluate: \( \int \frac{dx}{(e^x - 1)^2} \)
Answer: Let \( I = \int \frac{dx}{(e^x - 1)^2} dx \).

Put \( t = e^x - 1 \), so \( e^x = t + 1 \) and \( dt = e^x \, dx = (t + 1) dx \), giving \( dx = \frac{dt}{t + 1} \).

Then \( I = \int \frac{1}{t^2} \cdot \frac{dt}{t + 1} = \int \frac{dt}{t^2(t + 1)} \).

Using partial fractions, decompose \( \frac{1}{t^2(t + 1)} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t + 1} \).

This gives \( A \cdot t(t + 1) + B(t + 1) + C \cdot t^2 = 1 \).

Setting \( t = 0 \) yields \( B = 1 \).

Setting \( t = -1 \) yields \( C = 1 \).

Comparing coefficients of \( t^2 \): \( A + C = 0 \), so \( A = -1 \).

Therefore, \( \frac{1}{t^2(t + 1)} = -\frac{1}{t} + \frac{1}{t^2} + \frac{1}{t + 1} \).

\( I = -\int \frac{dt}{t} + \int \frac{dt}{t^2} + \int \frac{dt}{t + 1} = -\log|t| - \frac{1}{t} + \log|t + 1| + c \)

\( = \log\left|\frac{t + 1}{t}\right| - \frac{1}{t} + c = \log\left|\frac{e^x}{e^x - 1}\right| - \frac{1}{e^x - 1} + c \)

In simple words: Replace the exponential expression with a single variable to convert the integral to a rational form. Use partial fractions to split it into basic logarithmic and power terms, then integrate and back-substitute.

Exam Tip: Remember that repeated factors in the denominator require separate terms for each power - here \( t^2 \) generates both \( \frac{1}{t} \) and \( \frac{1}{t^2} \) terms.

Question 48. Evaluate: \( \int \frac{dx}{x(x^5 + 1)} \)
Answer: Let \( I = \int \frac{dx}{x(x^5 + 1)} dx \).

Put \( t = x^5 \), so \( dt = 5x^4 \, dx \).

Then \( I = \int \frac{x^4 \, dx}{x^5(x^5 + 1)} = \frac{1}{5}\int \frac{dt}{t(t + 1)} \).

Using partial fractions, decompose \( \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \).

This gives \( A(t + 1) + Bt = 1 \).

Setting \( t = 0 \) yields \( A = 1 \).

Setting \( t = -1 \) yields \( B = -1 \).

Therefore, \( \frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1} \).

\( I = \frac{1}{5}\left(\int \frac{dt}{t} - \int \frac{dt}{t + 1}\right) = \frac{1}{5}(\log|t| - \log|t + 1|) + c = \frac{1}{5}\log\left|\frac{t}{t + 1}\right| + c \)

\( = \frac{1}{5}\log\left|\frac{x^5}{x^5 + 1}\right| + c = \log|x| - \frac{1}{5}\log|x^5 + 1| + c \)

In simple words: Substitute the power to create a simpler rational function, decompose using partial fractions into two basic logarithmic terms, and integrate each separately.

Exam Tip: When the exponent in the denominator is large, always try a substitution that makes the highest power the new variable - this reduces complexity dramatically.

Question 30. Evaluate: \( \int \frac{(x^2 + 1)}{(x - 1)^2(x + 3)} dx \)
Answer: Let \( I = \int \frac{x^2 + 1}{(x - 1)^2(x + 3)} dx \).

Using partial fractions, decompose \( \frac{x^2 + 1}{(x - 1)^2(x + 3)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 3} \).

This gives \( A(x - 1)(x + 3) + B(x + 3) + C(x - 1)^2 = x^2 + 1 \).

Setting \( x = 1 \) yields \( B(4) = 2 \), so \( B = \frac{1}{2} \).

Setting \( x = -3 \) yields \( C(16) = 10 \), so \( C = \frac{5}{8} \).

Comparing coefficients of \( x^2 \): \( A + C = 1 \), so \( A = 1 - \frac{5}{8} = \frac{3}{8} \).

Therefore, \( \frac{x^2 + 1}{(x - 1)^2(x + 3)} = \frac{3}{8} \cdot \frac{1}{x - 1} + \frac{1}{2} \cdot \frac{1}{(x - 1)^2} + \frac{5}{8} \cdot \frac{1}{x + 3} \).

\( I = \frac{3}{8}\int \frac{dx}{x - 1} + \frac{1}{2}\int \frac{dx}{(x - 1)^2} + \frac{5}{8}\int \frac{dx}{x + 3} \)

\( = \frac{3}{8}\log|x - 1| - \frac{1}{2} \cdot \frac{1}{x - 1} + \frac{5}{8}\log|x + 3| + c \)

In simple words: Set up partial fractions with all powers of the repeated factor, find coefficients by substitution and comparison, then integrate using logarithmic and power formulas.

Exam Tip: Always organize the partial fraction setup before substitution - including all necessary terms prevents missing coefficients.

Question 31. Evaluate: \( \int \frac{x^2 + 1}{(x + 3)(x - 2)^2} dx \)
Answer: Let \( I = \int \frac{x^2 + 1}{(x - 3)(x - 1)^2} dx \)
Now setting \( \frac{x^2 + 1}{(x - 3)(x - 1)^2} = \frac{A}{(x - 3)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2} \) ... (1)
We get: \( A(x - 1)^2 + B(x - 3)(x - 1) + C(x - 3) = x^2 + 1 \)
When \( x - 1 = 0 \), so \( x = 1 \):
\( A(0) + B(0) + C(1 - 3) = 1 + 1 \)
\( C = -1 \)
When \( x - 3 = 0 \), so \( x = 3 \):
\( A(3 - 1)^2 + B(0) + C(0) = 9 + 1 \)
\( A(4) = 10 \)
\( A = \frac{5}{2} \)
Matching coefficients of \( x^2 \): \( A + B = 1 \)
\( \frac{5}{2} + B = 1 \)
\( B = 1 - \frac{5}{2} = -\frac{3}{2} \)
From equation (1), we get:
\( I = \int \frac{x^2 + 1}{(x - 3)(x - 1)^2} dx = \frac{5}{2} \int \frac{1}{x - 3} dx - \frac{3}{2} \int \frac{1}{x - 1} dx - \int \frac{1}{(x - 1)^2} dx \)
\( = \frac{5}{2} \log|x - 3| - \frac{3}{2} \log|x - 1| + \frac{1}{x - 1} + C \)
In simple words: Express the fraction as partial fractions using the decomposition method. Find constants A, B, and C by substituting convenient values and comparing coefficients. Then integrate each term separately to get the final result with logarithmic and reciprocal functions.

Exam Tip: Always verify partial fraction decomposition by recombining the fractions - this prevents algebraic errors. Remember that repeated linear factors require separate terms for each power.

Question 32. Evaluate: \( \int \frac{dx}{x(x^6 + 1)} \)
Answer: Let \( I = \int \frac{dx}{x(x^6 + 1)} \)
Set \( t = x^6 \), so \( dt = 6x^5 dx \)
\( I = \int \frac{dt}{6x^6(t + 1)} = \frac{1}{6} \int \frac{dt}{t(t + 1)} = \frac{1}{6} \int \frac{dt}{t(t + 1)} \)
Using partial fractions: \( \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \) ... (1)
We have: \( A(t + 1) + Bt = 1 \)
When \( t + 1 = 0 \), so \( t = -1 \):
\( A(0) + B(-1) = 1 \)
\( B = -1 \)
When \( t = 0 \):
\( A(0 + 1) + B(0) = 1 \)
\( A = 1 \)
\( \frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1} \)
\( \int \frac{1}{t(t + 1)} dt = \int \frac{1}{t} dt - \int \frac{1}{t + 1} dt = \log|t| - \log|t + 1| + c = \log \left| \frac{t}{t + 1} \right| + c \)
\( \int \frac{dx}{x(x^6 + 1)} = \frac{1}{6} \int \frac{dt}{t(t + 1)} = \frac{1}{6} \log \left| \frac{x^6}{x^6 + 1} \right| + c = \log x - \frac{1}{6} \log|x^6 + 1| + c \)
In simple words: Substitute to change the variable into a simpler form. Apply partial fraction decomposition to break down the expression. Integrate each fraction and substitute back the original variable.

Exam Tip: When dealing with substitutions, always track the differential change carefully - this ensures correct integral transformation. Use logarithmic properties to simplify your final answer.

Question 33. Evaluate: \( \int \frac{2x}{(2x + 1)^2} dx \)
Answer: Let \( I = \int \frac{2x}{(2x + 1)^2} dx \)
Using partial fractions: \( \frac{2x}{(2x + 1)^2} = \frac{A}{(2x + 1)} + \frac{B}{(2x + 1)^2} \) ... (1)
We get: \( A(2x + 1) + B = 2x \)
When \( 2x + 1 = 0 \):
\( A(0) + B = -1 \)
\( B = -1 \)
Matching coefficients of \( x \): \( 2A = 2 \), so \( A = 1 \)
From equation (1), we obtain:
\( \frac{2x}{(2x + 1)^2} = \frac{1}{(2x + 1)} - \frac{1}{(2x + 1)^2} \)
\( \int \frac{2x}{(2x + 1)^2} dx = \int \frac{1}{(2x + 1)} dx - \int \frac{1}{(2x + 1)^2} dx = \frac{\log|2x + 1|}{2} + \frac{1}{2(2x + 1)} + c \)
\( = \frac{1}{2} \left[ \log|2x + 1| + \frac{1}{2x + 1} \right] + c \)
In simple words: Break the fraction into simpler parts using partial fractions. Integrate each part - the first part gives a logarithm and the second part gives a reciprocal function.

Exam Tip: For repeated linear factors in the denominator, use separate terms for each power - a linear denominator and a squared denominator. Verify by adding the fractions back together.

Question 34. Evaluate: \( \int \frac{3x - 1}{(x + 2)(x - 2)^2} dx \)
Answer: Let \( I = \int \frac{3x + 1}{(x + 2)(x - 2)^2} dx \)
Using partial fractions: \( \frac{3x + 1}{(x + 2)(x - 2)^2} = \frac{A}{(x + 2)} + \frac{B}{(x - 2)} + \frac{C}{(x - 2)^2} \) ... (1)
We have: \( A(x - 2)^2 + B(x + 2)(x - 2) + C(x + 2) = 3x + 1 \)
When \( x - 2 = 0 \), so \( x = 2 \):
\( A(0) + B(0) + C(2 + 2) = 3 \times 2 + 1 \)
\( C = \frac{7}{4} \)
When \( x + 2 = 0 \), so \( x = -2 \):
\( A(-4)^2 + B(0) + C(0) = -6 + 1 = -5 \)
\( A = -\frac{5}{16} \)
Matching coefficients of \( x^2 \): \( A + B = 0 \)
\( B = \frac{5}{16} \)
From equation (1), we get:
\( I = -\frac{5}{16} \log|x + 2| + \frac{5}{16} \log|x - 2| - \frac{7}{4(x - 2)} + c \)
In simple words: Decompose the complex fraction into three simpler components. Find each constant by substituting strategic values and comparing coefficients. Integrate term by term to obtain the result.

Exam Tip: When the denominator has a repeated factor, create separate fractions for each power of that factor. Always check your decomposition by cross-multiplying and simplifying.

Question 35. Evaluate: \( \int \frac{5x + 8}{x^2(3x + 8)} dx \)
Answer: Let \( I = \int \frac{5x + 8}{x^2(3x + 8)} dx \)
Using partial fractions: \( \frac{5x + 8}{x^2(3x + 8)} = \frac{A}{(3x + 8)} + \frac{Bx + C}{x^2} \) ... (1)
We get: \( Ax^2 + (Bx + C)(3x + 8) = 5x + 8 \)
When \( 3x + 8 = 0 \), so \( x = -\frac{8}{3} \):
\( A\left(\frac{64}{9}\right) + B(0) = 5 \left(-\frac{8}{3}\right) + 8 \)
\( A = -\frac{3}{4} \)
Matching coefficients of \( x^2 \) and constant term:
\( A + 3B = 0 \), so \( B = \frac{1}{4} \)
\( 8C = 8 \), so \( C = 1 \)
From equation (1), we get:
\( I = \int \frac{5x + 8}{x^2(3x + 8)} dx = -\frac{3}{4} \times \frac{\log(3x + 8)}{3} + \frac{1}{4} \log x - \frac{1}{x} + c \)
\( = -\frac{1}{4} \log|3x + 8| + \frac{1}{4} \log x - \frac{1}{x} + c \)
In simple words: Split the fraction into partial terms with appropriate forms. Solve for the unknown coefficients using substitution and coefficient matching. Integrate each resulting term to obtain the final answer.

Exam Tip: For a fraction with \( x^2 \) in the denominator, include both a linear numerator term and a constant term in the decomposition. Always combine logarithmic results at the end using logarithm properties.

Question 36. Evaluate: \( \int \frac{5x^2 - 18x + 17}{(x - 1)^2(2x - 3)} dx \)
Answer: Let \( I = \int \frac{5x^2 - 18x + 17}{(x - 1)^2(2x - 3)} dx \)
Using partial fractions: \( \frac{5x^2 - 18x + 17}{(x - 1)^2(2x - 3)} = \frac{A}{(2x - 3)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2} \) ... (1)
We get: \( A(x - 1)^2 + B(2x - 3)(x - 1) + C(2x - 3) = 5x^2 - 18x + 17 \)
When \( x - 1 = 0 \), so \( x = 1 \):
\( A(0) + B(0) + C(2 - 3) = 5 - 18 + 17 \)
\( C(-1) = 4 \)
\( C = -4 \)
When \( 2x - 3 = 0 \), so \( x = \frac{3}{2} \):
\( A\left(\frac{1}{4}\right) + B(0) + C(0) = 5 \times \frac{9}{4} - 18 \times \frac{3}{2} + 17 \)
\( A = 5 \)
Matching coefficients of \( x^2 \): \( A + 2B = 5 \)
\( B = 0 \)
From equation (1), we get:
\( I = \int \frac{5x^2 - 18x + 17}{(x - 1)^2(2x - 3)} dx = 5 \times \frac{\log(2x - 3)}{2} + 0 - 4 \times \frac{1}{(x - 1)} + c \)
\( = \frac{5}{2} \log(2x - 3) + \frac{4}{x - 1} + c \)
In simple words: Decompose the rational function using partial fractions with three separate terms. Determine each constant by strategic substitution and coefficient comparison. Integrate each component to arrive at the solution.

Exam Tip: Handle repeated factors systematically - use one term for each power. Substitute the zero of each factor into the identity to isolate constants quickly and reduce computational errors.

Question 37. Evaluate: \( \int \frac{8}{(x + 2)(x^2 + 4)} dx \)
Answer: Let \( I = \int \frac{8}{(x + 2)(x^2 + 4)} dx \)
Using partial fractions: \( \frac{8}{(x + 2)(x^2 + 4)} = \frac{A}{(x + 2)} + \frac{Bx + C}{(x^2 + 4)} \) ... (1)
We get: \( A(x^2 + 4) + (Bx + C)(x + 2) = 8 \)
When \( x + 2 = 0 \), so \( x = -2 \):
\( A(4 + 4) + 0 = 8 \)
\( A = 1 \)
Matching coefficients of \( x^2 \) and constant term:
\( A + B = 0 \), so \( B = -1 \)
\( 4A + 2C = 8 \), so \( C = 2 \)
From equation (1), we obtain:
\( I = \int \frac{8}{(x + 2)(x^2 + 4)} dx = \int \frac{1}{(x + 2)} dx - \int \frac{x}{(x^2 + 4)} dx + 2 \int \frac{1}{(x^2 + 4)} dx \)
\( = \log|x + 2| - \frac{1}{2} \log(x^2 + 4) + 2 \times \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + c \)
\( = \log|x + 2| - \frac{1}{2} \log|x^2 + 4| + \tan^{-1}\left(\frac{x}{2}\right) + c \)
In simple words: Write the fraction as a sum of three partial terms. Solve for constants using substitution and comparison. Integrate - the first term gives a logarithm, the second a logarithm, and the third an inverse tangent function.

Exam Tip: When the denominator includes an irreducible quadratic, use a linear numerator (Bx + C) in that partial fraction. Recognize the standard integral \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).

Question 38. Evaluate: \( \int \frac{3x + 5}{(x^3 - x^2 - x - 1)} dx \)
Answer: Let \( I = \int \frac{3x + 5}{(x^3 - x^2 - x - 1)} dx \)
Using partial fractions: \( \frac{3x + 5}{(x^3 - x^2 - x - 1)} = \frac{A}{(x - 1)} + \frac{Bx + C}{(x^2 + 1)} \) ... (1)
We get: \( A(x^2 + 1) + (Bx + C)(x - 1) = 3x + 5 \)
When \( x - 1 = 0 \), so \( x = 1 \):
\( A(2) + B(0) = 3 + 5 = 8 \)
\( A = 4 \)
Matching coefficients of \( x^2 \) and constant term:
\( A + B = 0 \), so \( B = -4 \)
\( A - C = 5 \), so \( C = -1 \)
From equation (1), we get:
\( I = \int \frac{3x + 5}{(x - 1)(x^2 + 1)} dx = 4 \int \frac{1}{(x - 1)} dx - 4 \int \frac{x}{(x^2 + 1)} dx - \int \frac{1}{(x^2 + 1)} dx \)
\( = 4 \log(x - 1) - 2 \log(x^2 + 1) - \tan^{-1} x + c \)
In simple words: Express the function as a combination of partial fractions. Find the constants A, B, and C using algebraic methods. Integrate each term separately to obtain logarithmic and inverse tangent components.

Exam Tip: Always factor the denominator completely before setting up partial fractions. Irreducible quadratic factors should have a linear numerator; use this standard form consistently.

Question 49. Evaluate: \( \int \frac{dx}{x(x + 1)} \)
Answer: Let \( I = \int \frac{dx}{x(x + 1)} \)
Using partial fractions: \( \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{(x + 1)} \) ... (1)
We have: \( A(x + 1) + Bx = 1 \)
When \( x + 1 = 0 \), so \( x = -1 \):
\( A(0) + B(-1) = 1 \)
\( B = -1 \)
When \( x = 0 \):
\( A(0 + 1) + B(0) = 1 \)
\( A = 1 \)
\( \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{(x + 1)} \)
\( I = \int \frac{1}{x(x + 1)} dt = \int \frac{1}{x} dt - \int \frac{1}{(x + 1)} dt = \log t - \log|t + 1| + c \)
\( = \log \left| \frac{x}{x + 1} \right| + c \)
In simple words: Break the fraction into two simpler fractions using partial decomposition. Each fraction integrates to a logarithm. Combine the logarithmic results using logarithm laws.

Exam Tip: For products of distinct linear factors, use one partial fraction term for each factor. The constants can be found quickly by substituting the zeros of each factor.

Question 50. Evaluate: \( \int \frac{dx}{\sin x (3 + 2\cos x)} \)
Answer: Let \( I = \int \frac{dx}{\sin x (3 + 2\cos x)} \)
Substitute \( t = \cos x \), so \( dt = -\sin x dx \)
\( I = \int \frac{dt}{-\sin x \times \sin x (3 + 2t)} = -\int \frac{dt}{\sin^2 x (3 + 2t)} = -\int \frac{dt}{(1 - \cos^2 x)(3 + 2t)} = -\int \frac{dt}{(1 - t^2)(3 + 2t)} \)
Using partial fractions: \( \frac{1}{(1 - t^2)(3 + 2t)} = \frac{1}{(1 - t)(1 + t)(3 + 2t)} = \frac{A}{(1 - t)} + \frac{B}{(1 + t)} + \frac{C}{(3 + 2t)} \) ... (1)
We get: \( A(1 + t)(3 + 2t) + B(1 - t)(3 + 2t) + C(1 + t)(1 - t) = 1 \)
When \( 1 + t = 0 \), so \( t = -1 \):
\( A(0) + B(2)(3 - 2) + C(0) = 1 \)
\( B = \frac{1}{2} \)
When \( 1 - t = 0 \), so \( t = 1 \):
\( A(2)(5) + B(0) + C(0) = 1 \)
\( A = \frac{1}{10} \)
When \( 3 + 2t = 0 \), so \( t = -\frac{3}{2} \):
\( A(0) + B(0) + C\left(1 - \frac{9}{4}\right) = 1 \)
\( C = -\frac{4}{5} \)
\( I = -\int \frac{dt}{(1 - t^2)(3 + 2t)} = -\left[\frac{1}{10} \int \frac{1}{(1 - t)} dt + \frac{1}{2} \int \frac{1}{(1 + t)} dt - \frac{4}{5} \int \frac{1}{(3 + 2t)} dt\right] \)
\( = -\frac{1}{10} \log|1 - t| + \frac{1}{2} \log|1 + t| - \frac{2}{5} \log|3 + 2t| + c \)
\( = -\frac{1}{10} \log|1 - \cos x| + \frac{1}{2} \log|1 + \cos x| - \frac{2}{5} \log|3 + 2\cos x| + c \)
In simple words: Use substitution to simplify the trigonometric expression into a rational function. Decompose using partial fractions into three simpler terms. Integrate each term and back-substitute the original variable.

Exam Tip: When working with trigonometric integrands, choose substitutions that convert sine and cosine into a rational variable. Always account for the differential carefully during substitution.

Question 51. Evaluate: \( \int \frac{dx}{\cos x (5 - 4 \sin x)} \)
Answer: Let \( I = \int \frac{dx}{\cos x (5 - 4 \sin x)} \)
Substitute \( t = \sin x \), so \( dt = \cos x dx \)
\( I = \int \frac{dt}{(1 - \sin^2 x)(5 - 4t)} = \int \frac{dt}{(1 - t^2)(5 - 4t)} \)
Using partial fractions: \( \frac{1}{(1 - t^2)(5 - 4t)} = \frac{1}{(1 - t)(1 + t)(5 - 4t)} = \frac{A}{(1 - t)} + \frac{B}{(1 + t)} + \frac{C}{(5 - 4t)} \) ... (1)
We get: \( A(1 + t)(5 - 4t) + B(1 - t)(5 - 4t) + C(1 + t)(1 - t) = 1 \)
When \( 1 + t = 0 \), so \( t = -1 \):
\( A(0) + B(2)(9) + C(0) = 1 \)
\( B = \frac{1}{18} \)
When \( 1 - t = 0 \), so \( t = 1 \):
\( A(2) + B(0) + C(0) = 1 \)
\( A = \frac{1}{2} \)
When \( 5 - 4t = 0 \), so \( t = \frac{5}{4} \):
\( A(0) + B(0) + C\left(1 - \frac{25}{16}\right) = 1 \)
\( C = -\frac{16}{9} \)
\( I = \int \frac{dt}{(1 - t^2)(5 - 4t)} = -\frac{1}{2} \log|1 - t| + \frac{1}{18} \log|1 + t| + \frac{4}{9} \log|5 - 4\sin x| + c \)
In simple words: Convert the trigonometric integral to a rational one through substitution. Use partial fraction decomposition for the resulting rational expression. Integrate each fraction and substitute back to express in terms of the original variable.

Exam Tip: Choose substitutions that align with the structure of the trigonometric functions in the denominator. Verify your constants by substituting them back into the decomposition equation.

Question 52. Evaluate: \( \int \frac{dx}{\sin x \cos^2 x} \)
Answer: Let \( I = \int \frac{dx}{\sin x \cos^2 x} = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^2 x} dx = \int \frac{\sin x}{\sin x \cos^2 x} dx + \int \frac{\cos^2 x}{\sin x \cos^2 x} dx \)
\( = \int \frac{\sin x}{\cos^2 x} dx + \int \frac{1}{\sin x} dx = \int (\tan x \sec x + \csc x) dx \)
\( = \sec x - \frac{1}{2} \log \cot^2 \frac{x}{2} = \sec x - \frac{1}{2} \log \left( \frac{1 + \cos x}{1 - \cos x} \right) + c \)
In simple words: Rewrite the integrand by adding and subtracting squared terms in the numerator. Split into two integrals. Each resulting integral is a standard form - one yields a secant function and the other yields a logarithmic expression.

Exam Tip: Recognizing standard trigonometric integral forms \( \int \tan x \sec x dx \) and \( \int \csc x dx \) is essential. Use half-angle formulas to express logarithmic results in alternate forms if needed.

Question 53. Evaluate: \( \int \frac{\tan x}{(1 - \sin x)} dx \)
Answer: Let \( I = \int \frac{\tan x}{(1 - \sin x)} dx = \int \frac{\sin x}{\cos x (1 - \sin x)} dx \)
Substitute \( t = \sin x \), so \( dt = \cos x dx \)
\( I = \int \frac{\sin x \times \cos x}{\cos^2 x (1 - \sin x)} dx = \int \frac{tdt}{\cos^2 x (1 - t)} = \int \frac{tdt}{(1 - \sin^2 x)(1 - t)} = \int \frac{tdt}{(1 - t^2)(1 - t)} \)
Using partial fractions: \( \frac{t}{(1 - t)(1 + t)(1 - t)} = \frac{A}{(1 + t)} + \frac{B}{(1 - t)} + \frac{C}{(1 - t)^2} \) ... (1)
We get: \( A(1 + t)^2 + B(1 - t)(1 + t) + C(1 + t) = t \)
When \( 1 - t = 0 \), so \( t = 1 \):
\( A(0) + B(0) + C(1 + 1) = 1 \)
\( C = \frac{1}{2} \)
When \( 1 + t = 0 \), so \( t = -1 \):
\( A(2)^2 + B(0) + C(0) = -1 \)
\( A = -\frac{1}{4} \)
Matching coefficients of \( t^2 \): \( A - B = 0 \), so \( B = -\frac{1}{4} \)
\( I = \int \frac{tdt}{(1 - t)(1 + t)(1 - t)} = -\frac{1}{4} \int \frac{1}{(1 + t)} dt - \frac{1}{4} \int \frac{1}{(1 - t)} dt + \frac{1}{2} \int \frac{1}{(1 - t)^2} dt \)
\( = -\frac{1}{4} \log|1 + t| - \frac{1}{4} \log|1 - t| - \frac{1}{2} \times \frac{1}{(1 - t)} + c \)
\( = -\frac{1}{4} \log|1 + \sin x| - \frac{1}{4} \log|1 - \sin x| - \frac{1}{2} \times \frac{1}{(1 - \sin x)} + c \)
In simple words: Use substitution to transform the trigonometric integral into a rational function. Decompose the rational expression into partial fractions with three terms. Integrate each separately and back-substitute.

Exam Tip: For repeated linear factors in the denominator after substitution, include terms for each power. Always simplify logarithmic products - combine them into a single logarithm when possible.

Question 54. Evaluate: \( \int \frac{dx}{\sin x + \sin 2x} \)
Answer: Let \( I = \int \frac{dx}{\sin x + \sin 2x} = \int \frac{dx}{(\sin x + 2 \sin x \cos x)} = \int \frac{dx}{\sin x (1 + 2 \cos x)} \)
Substitute \( t = \cos x \), so \( dt = -\sin x dx \)
\( I = -\int \frac{dt}{\sin^2 x (1 + 2t)} = -\int \frac{dt}{(1 - \cos^2 x)(1 + 2t)} = -\int \frac{dt}{(1 - t^2)(1 + 2t)} \)
Using partial fractions: \( \frac{1}{(1 - t^2)(1 + 2t)} = \frac{A}{(1 - t)} + \frac{B}{(1 + t)} + \frac{C}{(1 + 2t)} \) ... (1)
We get: \( A(1 + t)(1 + 2t) + B(1 - t)(1 + 2t) + C(1 - t^2) = 1 \)
When \( 1 + t = 0 \), so \( t = -1 \):
\( A(0) + B(2)(-1 - 2) + C(0) = 1 \)
\( B = -\frac{1}{2} \)
When \( 1 - t = 0 \), so \( t = 1 \):
\( A(2)(3) + B(0) + C(0) = 1 \)
\( A = \frac{1}{6} \)
When \( 1 + 2t = 0 \), so \( t = -\frac{1}{2} \):
\( A(0) + B(0) + C\left(1 - \frac{1}{4}\right) = 1 \)
\( C = \frac{4}{3} \)
\( I = -\left[\frac{1}{6} \int \frac{1}{(1 - t)} dt - \frac{1}{2} \int \frac{1}{(1 + t)} dt + \frac{4}{3} \int \frac{1}{(1 + 2t)} dt\right] \)
\( = \frac{1}{6} \log|1 - t| + \frac{1}{2} \log|1 + t| - \frac{2}{3} \log|1 + 2t| + c \)
\( = \frac{1}{6} \log|1 - \cos x| + \frac{1}{2} \log|1 + \cos x| - \frac{2}{3} \log|1 + 2\cos x| + c \)
In simple words: Factor the denominator using trigonometric identities. Substitute to convert to a rational form. Decompose and integrate using partial fractions, then substitute back the original variable.

Exam Tip: Use double-angle formulas to simplify trigonometric expressions before integrating. The substitution \( t = \cos x \) or \( t = \sin x \) often converts these integrals into manageable rational forms.

Question 55. Evaluate: \( \int \frac{x^2}{(x^4 - x^2 - 12)} dx \)
Answer: Let \( I = \int \frac{x^2}{(x^4 - x^2 - 12)} dx \)
Substitute \( t = x^2 \), so the denominator becomes \( t^2 - t - 12 = (t - 4)(t + 3) \)
Using partial fractions: \( \frac{t}{(t - 4)(t + 3)} = \frac{A}{(t - 4)} + \frac{B}{(t + 3)} \)
We get: \( A(t + 3) + B(t - 4) = t \)
When \( t + 3 = 0 \), so \( t = -3 \):
\( A(0) + B(-7) = -3 \)
\( B = \frac{3}{7} \)
When \( t - 4 = 0 \), so \( t = 4 \):
\( A(4 + 3) + B(0) = 4 \)
\( A = \frac{4}{7} \)
From equation (1):
\( \frac{t}{(t - 4)(t + 3)} = \frac{4}{7} \times \frac{1}{(t - 4)} + \frac{3}{7} \times \frac{1}{(t + 3)} \)
\( \frac{x^2}{(x^2 - 4)(x^2 + 3)} = \frac{4}{7} \times \frac{1}{(x^2 - 2^2)} + \frac{3}{7} \times \frac{1}{(x^2 + (\sqrt{3})^2)} \)
\( \int \frac{x^2}{(x^4 - x^2 - 12)} dx = \frac{4}{7} \int \frac{1}{(x^2 - 4)} dx + \frac{3}{7} \int \frac{1}{(x^2 + 3)} dx \)
\( = \frac{4}{7} \times \frac{1}{2 \times 2} \log \left| \frac{x - 2}{x + 2} \right| + \frac{3}{7} \times \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + c \)
\( = \frac{1}{7} \log \left| \frac{x - 2}{x + 2} \right| + \frac{\sqrt{3}}{7} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + c \)
In simple words: Substitute to convert the quartic expression into a quadratic in the new variable. Use partial fractions on the resulting rational expression. Integrate each term using standard formulas for difference of squares and sum of squares.

Exam Tip: When dealing with \( x^4 \) terms, substituting \( t = x^2 \) typically reduces the problem to a manageable degree. Always factor the resulting quadratic completely to identify the partial fraction form.

Question 56. Evaluate: \( \int \frac{x^4}{(x^2 + 1)(x^2 + 9)(x^2 + 16)} dx \)
Answer: Let \( I = \int \frac{x^4}{(x^2 + 1)(x^2 + 9)(x^2 + 16)} dx \)
Substitute \( t = x^2 \), so the denominator becomes \( (t + 1)(t + 9)(t + 16) \)
Using partial fractions: \( \frac{t^2}{(t + 1)(t + 9)(t + 16)} = \frac{A}{(t + 1)} + \frac{B}{(t + 9)} + \frac{C}{(t + 16)} \) ... (1)
We get: \( t^2 = A(t + 9)(t + 16) + B(t + 1)(t + 16) + C(t + 1)(t + 9) \)
When \( t + 1 = 0 \), so \( t = -1 \):
\( A(8)(15) + B(0) + C(0) = 1 \)
\( A = \frac{1}{120} \)
When \( t + 9 = 0 \), so \( t = -9 \):
\( A(0) + B(-8)(7) + C(0) = 81 \)
\( B = -\frac{81}{56} \)
When \( t + 16 = 0 \), so \( t = -16 \):
\( A(0) + B(0) + C(-15)(-7) = 256 \)
\( C = \frac{256}{105} \)
From equation (1):
\( \frac{t^2}{(t + 1)(t + 9)(t + 16)} = \frac{1}{120} \times \frac{1}{(t + 1)} - \frac{81}{56} \times \frac{1}{(t + 9)} + \frac{256}{105} \times \frac{1}{(t + 16)} \)
\( \int \frac{x^4}{(x^2 + 1)(x^2 + 9)(x^2 + 16)} dx = \frac{1}{120} \int \frac{1}{(x^2 + 1)} dx - \frac{81}{56} \int \frac{1}{(x^2 + 9)} dx + \frac{256}{105} \int \frac{1}{(x^2 + 16)} dx \)
\( = \frac{1}{120} \tan^{-1} x - \frac{81}{56} \times \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) + \frac{256}{105} \times \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + c \)
\( = \frac{1}{120} \tan^{-1} x - \frac{27}{56} \tan^{-1} \left( \frac{x}{3} \right) + \frac{64}{105} \tan^{-1} \left( \frac{x}{4} \right) + c \)
In simple words: Substitute \( t = x^2 \) to transform the integral. Apply partial fractions to the resulting polynomial divided by three linear factors. Integrate each resulting term using the inverse tangent integral formula.

Exam Tip: For products of irreducible quadratic factors, the substitution \( t = x^2 \) is powerful. Remember that \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \) is a key standard formula.

Question 39. Evaluate: \( \int \frac{2x}{(x^2+1)(x^2-3)} dx \)
Answer: Let \( I = \int \frac{2x}{(x^2+1)(x^2+3)} dx \). Set \( t = x^2 \), so \( dt = 2x \, dx \). Now, decompose using partial fractions:
\( \frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3} \)
This gives \( A(t+3) + B(t+1) = 1 \). Setting \( t = -3 \) yields \( B = -\frac{1}{2} \); setting \( t = -1 \) yields \( A = \frac{1}{2} \). Therefore:
\( \frac{1}{(t+1)(t+3)} = \frac{1}{2} \cdot \frac{1}{t+1} - \frac{1}{2} \cdot \frac{1}{t+3} \)
\( \int \frac{1}{(t+1)(t+3)} dt = \frac{1}{2} \log|t+1| - \frac{1}{2} \log|t+3| + c \)
\( \int \frac{2x}{(x^2+1)(x^2+3)} dx = \frac{1}{2} \log|x^2+1| - \frac{1}{2} \log|x^2+3| + c \)

Exam Tip: Verify partial fraction coefficients by substitution before integrating - this prevents sign errors in logarithmic answers.

Question 40. Evaluate: \( \int \frac{x^2}{(x^2-1)} dx \)
Answer: Let \( I = \int \frac{x^2}{(x^2-1)} dx \). Set \( t = x^2 \), so \( dt = 2x \, dx \). Using partial fractions on \( \frac{t}{(t-1)(t+1)} \), we decompose as:
\( \frac{t}{(t-1)(t+1)} = \frac{A}{t-1} + \frac{B}{t+1} \)
This gives \( A(t+1) + B(t-1) = t \). Setting \( t = 1 \) yields \( A = \frac{1}{2} \); setting \( t = -1 \) yields \( B = \frac{1}{2} \). Therefore:
\( \frac{t}{(t-1)(t+1)} = \frac{1}{2} \cdot \frac{1}{t-1} + \frac{1}{2} \cdot \frac{1}{t+1} \)
\( \int \frac{x^2}{(x^4-1)} dt = \frac{1}{2} \int \frac{1}{x^2-1} dt + \frac{1}{2} \int \frac{1}{x^2+1} dt \)
\( = \frac{1}{4} \log \left|\frac{x-1}{x+1}\right| + \frac{1}{2} \tan^{-1} x + c \)

Exam Tip: Break improper fractions using long division before applying partial fractions to simplify the decomposition.

Question 41. Evaluate: \( \int \frac{2}{(1-x)(1+x^2)} dx \)
Answer: Let \( I = \int \frac{2}{(1-x)(1+x^2)} dx \). Using partial fractions:
\( \frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{x^2+1} \)
Multiplying both sides by \( (1-x)(1+x^2) \) gives \( A(1+x^2) + (Bx+C)(1-x) = 2 \). Substituting \( x = 1 \) yields \( A = 1 \); substituting \( x = 0 \) yields \( C = 1 \); comparing coefficients of \( x \) gives \( B = 1 \). Therefore:
\( \frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2} \)
\( \int \frac{2}{(1-x)(1+x^2)} dx = -\log|1-x| + \frac{1}{2} \log|1+x^2| + \tan^{-1} x + c \)

Exam Tip: When a quadratic factor appears in the denominator, always use a linear numerator \( Bx+C \) for that part.

Question 42. Evaluate: \( \int \frac{2x^2+1}{x^2(x^2+4)} dx \)
Answer: Let \( I = \int \frac{2x^2+1}{x^2(x^2+4)} dx \). Using partial fractions:
\( \frac{2x^2+1}{x^2(x^2+4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+4} \)
Multiplying by \( x^2(x^2+4) \) gives \( A \cdot x(x^2+4) + B(x^2+4) + (Cx+D) x^2 = 2x^2+1 \). Setting \( x = 0 \) yields \( B = \frac{1}{4} \); expanding and comparing coefficients determines the remaining constants. The integral evaluates to:
\( \int \frac{2x^2+1}{x^2(x^2+4)} dx = -\frac{1}{4x} + \frac{7}{8} \tan^{-1}\left(\frac{x}{2}\right) + c \)

Exam Tip: For repeated linear factors, assign a separate constant to each power; for irreducible quadratics, use a linear expression in the numerator.

Question 43. Evaluate: \( \int \sin 3x \, dx \)
Answer: To find \( \int \sin 3x \, dx \), apply the standard formula \( \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + c \). With \( a = 3 \):
\( \int \sin 3x \, dx = -\frac{1}{3} \cos 3x + c \)

Exam Tip: Remember to divide by the coefficient inside the sine function, not multiply.

Question 44. Evaluate: \( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx \)
Answer: Rewrite the integrand as \( \int \left(x^{1/2} + x^{-1/2}\right) dx \). Using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + c \):
\( \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \)
\( \int \frac{1}{\sqrt{x}} dx = \frac{x^{1/2}}{1/2} = 2x^{1/2} \)
\( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = \frac{2}{3} x^{3/2} + 2\sqrt{x} + c \)

Exam Tip: Convert roots and reciprocals to exponential form first to apply the power rule correctly.

Question 45. Evaluate: \( \int \frac{x^2}{1+x^3} dx \)
Answer: Let \( t = 1 + x^3 \), so \( dt = 3x^2 \, dx \) or \( x^2 \, dx = \frac{1}{3} dt \). Substituting:
\( \int \frac{x^2}{1+x^3} dx = \int \frac{1}{t} \cdot \frac{1}{3} dt = \frac{1}{3} \ln|t| + c = \frac{1}{3} \ln(x^3+1) + c \)

Exam Tip: Recognize when the derivative of the denominator (or a close form) appears in the numerator - this signals a logarithmic integral.

Question 46. Evaluate: \( \int \frac{2\cos x}{3\sin^2 x} dx \)
Answer: Let \( t = \sin x \), so \( dt = \cos x \, dx \). Substituting:
\( \int \frac{2\cos x}{3\sin^2 x} dx = \int \frac{2}{3t^2} dt = \frac{2}{3} \cdot \left(-\frac{1}{t}\right) + c = -\frac{2}{3\sin x} + c \)

Exam Tip: When a trigonometric function and its derivative both appear, use substitution with the function itself.

Question 47. Evaluate: \( \int \frac{(3\sin\phi-2)\cos\phi}{(5-\cos^2\phi-4\sin\phi)} d\phi \)
Answer: First, simplify the denominator. Note that \( 5 - \cos^2\phi - 4\sin\phi = 5 - (1-\sin^2\phi) - 4\sin\phi = 4 + \sin^2\phi - 4\sin\phi = (\sin\phi-2)^2 \). Decompose the numerator:
\( \frac{3\cos\phi}{(\sin\phi-2)} + \frac{4\cos\phi}{(\sin\phi-2)^2} \)
Let \( t = \sin\phi - 2 \), so \( dt = \cos\phi \, d\phi \). Then:
\( \int \frac{3dt}{t} + \int \frac{4dt}{t^2} = 3\ln|t| - \frac{4}{t} + c = 3\ln|\sin\phi-2| - \frac{4}{\sin\phi-2} + c \)

Exam Tip: Factor and simplify the denominator before attempting decomposition - this often reveals a structure that makes substitution clearer.

Exercise 15B

Question 1. Evaluate: \( \int x^{-6} dx \)
Answer: Using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + c \) with \( n = -6 \):
\( \int x^{-6} dx = \frac{x^{-5}}{-5} + c = -\frac{1}{5x^5} + c \)

Exam Tip: Apply the power rule to negative exponents the same way - always add 1 to the exponent and divide by the new exponent.

Question 2. Evaluate: \( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx \)
Answer: Rewrite as \( \int \left(x^{1/2} + x^{-1/2}\right) dx \). Applying the power rule to each term:
\( \int x^{1/2} dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \)
\( \int x^{-1/2} dx = \frac{x^{1/2}}{1/2} = 2x^{1/2} \)
\( \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = \frac{2}{3} x^{3/2} + 2\sqrt{x} + c \)

Exam Tip: Convert radicals to fractional powers before integrating - the power rule works uniformly for all real exponents except -1.

Question 3. Evaluate: \( \int \sin 3x \, dx \)
Answer: Using the standard formula \( \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + c \) with \( a = 3 \):
\( \int \sin 3x \, dx = -\frac{1}{3} \cos 3x + c \)

Exam Tip: Dividing by the coefficient of \( x \) inside the function is essential - forgetting this step is a common error.

Question 4. Evaluate: \( \int \frac{x^2}{(1+x^3)} dx \)
Answer: Let \( t = 1 + x^3 \), so \( dt = 3x^2 \, dx \) or \( x^2 \, dx = \frac{1}{3} dt \). Substituting:
\( \int \frac{x^2}{1+x^3} dx = \int \frac{1}{t} \cdot \frac{1}{3} dt = \frac{1}{3} \ln|t| + c = \frac{1}{3} \ln(1+x^3) + c \)

Exam Tip: Recognize derivative patterns - when the numerator is (or is proportional to) the derivative of something in the denominator, a logarithmic result follows.

Question 5. Evaluate: \( \int \frac{2\cos x}{3\sin^2 x} dx \)
Answer: Let \( t = \sin x \), so \( dt = \cos x \, dx \). Rewriting:
\( \int \frac{2\cos x}{3\sin^2 x} dx = \int \frac{2}{3t^2} dt = \frac{2}{3} \int t^{-2} dt = \frac{2}{3} \cdot (-t^{-1}) + c = -\frac{2}{3\sin x} + c \)
In simple words: Substitute the trig function so the integral becomes a simple power, then use the power rule and substitute back.

Exam Tip: Choose substitutions where both the function and its derivative appear in the integrand.

Question 6. Evaluate: \( \int \frac{(3\sin\phi-2)\cos\phi}{(5-\cos^2\phi-4\sin\phi)} d\phi \)
Answer: Recognize that the denominator simplifies: \( 5 - \cos^2\phi - 4\sin\phi = 5 - (1-\sin^2\phi) - 4\sin\phi = (\sin\phi-2)^2 \). Split the numerator:
\( \frac{(3\sin\phi-2)\cos\phi}{(\sin\phi-2)^2} = \frac{3\cos\phi}{(\sin\phi-2)} + \frac{4\cos\phi}{(\sin\phi-2)^2} \)
With \( t = \sin\phi - 2 \) and \( dt = \cos\phi \, d\phi \):
\( \int \frac{3}{t} dt + \int \frac{4}{t^2} dt = 3\ln|t| - \frac{4}{t} + c = 3\ln|\sin\phi-2| - \frac{4}{\sin\phi-2} + c \)
In simple words: Simplify the denominator first, split the fraction into manageable pieces, then substitute to turn it into basic power and logarithm integrals.

Exam Tip: Always check if the denominator factors or simplifies - this often transforms a complex integral into one you can handle with standard substitution.

Question 7. Evaluate: \( \int \sin^2 x \, dx \)
Answer: Use the identity \( 1 - \cos 2x = 2\sin^2 x \), so \( \sin^2 x = \frac{1 - \cos 2x}{2} \). Then:
\( \int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} dx = \frac{1}{2} \int (1 - \cos 2x) dx = \frac{1}{2} \left( x - \frac{\sin 2x}{2} \right) + c = \frac{x}{2} - \frac{\sin 2x}{4} + c \)
In simple words: Replace the squared trig function with a half-angle identity, then integrate the simpler expression.

Exam Tip: For integrals of squared trig functions, use power-reduction formulas - they convert the problem into a sum of simple terms.

Question 8. Evaluate: \( \int \frac{(\log x)^2}{x} dx \)
Answer: Let \( t = \log x \), so \( \frac{1}{x} dx = dt \). Substituting:
\( \int \frac{(\log x)^2}{x} dx = \int t^2 \, dt = \frac{t^3}{3} + c = \frac{(\log x)^3}{3} + c \)
In simple words: When the logarithm and the differential \( \frac{dx}{x} \) both appear, substitute the logarithm itself to turn the integral into a power function.

Exam Tip: The differential \( \frac{dx}{x} \) is the derivative of \( \log x \) - recognize this pattern for smooth substitution.

Question 9. Evaluate: \( \int \frac{(x+1)(x + \log x)^2}{x} dx \)
Answer: Rewrite the integrand as \( \int \left(1 + \frac{1}{x}\right)(x + \log x)^2 dx \). Let \( t = x + \log x \), so \( dt = \left(1 + \frac{1}{x}\right) dx \). Substituting:
\( \int (x + \log x)^2 \left(1 + \frac{1}{x}\right) dx = \int t^2 \, dt = \frac{t^3}{3} + c = \frac{(x + \log x)^3}{3} + c \)
In simple words: Notice that the derivative of \( x + \log x \) is already present in the integrand - this makes substitution straightforward.

Exam Tip: Check if the derivative of part of the integrand appears elsewhere - this signals a substitution that will simplify the problem.

Question 10. Evaluate: \( \int \frac{\sin x}{(1+\cos x)} dx \)
Answer: Let \( t = 1 + \cos x \), so \( dt = -\sin x \, dx \). Substituting:
\( \int \frac{\sin x}{1+\cos x} dx = \int \frac{-dt}{t} = -\ln|t| + c = -\ln|1+\cos x| + c \)
In simple words: The sine in the numerator is (up to a sign) the derivative of cosine, which appears in the denominator - use this to set up your substitution.

Exam Tip: Always check the sign - a negative differential might appear, but that just changes the sign of the logarithm.

Question 11. Evaluate: \( \int \frac{(1+\tan x)}{(1-\tan x)} dx \)
Answer: Note that \( \frac{1+\tan x}{1-\tan x} = \frac{\cos x + \sin x}{\cos x - \sin x} \). Let \( t = \cos x - \sin x \), so \( dt = -(\sin x + \cos x) dx \). Then:
\( \int \frac{\cos x + \sin x}{\cos x - \sin x} dx = \int \frac{-dt}{t} = -\ln|t| + c = -\ln|\cos x - \sin x| + c \)
In simple words: Convert tangent to sines and cosines, recognize that the numerator is the negative derivative of the denominator, and integrate as a logarithm.

Exam Tip: When working with tangent or cotangent, rewrite in terms of sine and cosine first - this often reveals hidden derivative patterns.

Question 12. Evaluate: \( \int \frac{(1-\cot x)}{(1+\cot x)} dx \)
Answer: Express as \( \frac{1-\cot x}{1+\cot x} = \frac{\sin x - \cos x}{\sin x + \cos x} \). Let \( t = \sin x + \cos x \), so \( dt = (\cos x - \sin x) dx \). Then:
\( \int \frac{\sin x - \cos x}{\sin x + \cos x} dx = \int \frac{-dt}{t} = -\ln|t| + c = -\ln|\sin x + \cos x| + c \)
In simple words: Rewrite cotangent in terms of sine and cosine, match the numerator to the derivative of the denominator (with a sign), and integrate as a logarithm.

Exam Tip: For composite trig functions, always express them in basic sine and cosine form - this makes derivative matching visible.

Question 13. Evaluate: \( \int \frac{(1+\cot x)}{(x+\log \sin x)} dx \)
Answer: Let \( t = x + \log(\sin x) \), so \( dt = \left(1 + \cot x\right) dx \). Substituting:
\( \int \frac{1+\cot x}{x+\log \sin x} dx = \int \frac{dt}{t} = \ln|t| + c = \ln|x + \log \sin x| + c \)
In simple words: The derivative of \( x + \log(\sin x) \) is \( 1 + \cot x \), which is exactly your numerator - so substitute and integrate directly.

Exam Tip: Before diving into a complex integral, verify whether the numerator matches the derivative of the denominator - this saves time.

Question 14. Evaluate: \( \int \frac{(1-\sin 2x)}{(x+\cos^2 x)} dx \)
Answer: Let \( t = x + \cos^2 x \), so \( dt = (1 - \sin 2x) dx \). (Note: \( \frac{d}{dx}(\cos^2 x) = -2\cos x \sin x = -\sin 2x \).) Substituting:
\( \int \frac{1-\sin 2x}{x+\cos^2 x} dx = \int \frac{dt}{t} = \ln|t| + c = \ln|x + \cos^2 x| + c \)
In simple words: Recognize that the numerator is the derivative of the expression in the denominator, making this a standard logarithmic integral.

Exam Tip: Pay attention to double-angle formulas - \( \sin 2x \) and \( \cos 2x \) often appear as derivatives of composite expressions.

Question 15. Evaluate: \( \int \frac{\sec^2(\log x)}{x} dx \)
Answer: Let \( t = \log x \), so \( \frac{1}{x} dx = dt \). Substituting:
\( \int \frac{\sec^2(\log x)}{x} dx = \int \sec^2 t \, dt = \tan t + c = \tan(\log x) + c \)
In simple words: The differential \( \frac{dx}{x} \) is the differential of \( \log x \) - substitute the logarithm to turn this into a basic secant-squared integral.

Exam Tip: Recall that \( \int \sec^2 u \, du = \tan u + c \) - this is one of the fundamental integral formulas to memorize.

Question 16. Evaluate: \( \int \frac{\sin(2\tan^{-1} x)}{(1+x^2)} dx \)
Answer: Let \( t = \tan^{-1} x \), so \( \frac{1}{1+x^2} dx = dt \). Substituting:
\( \int \frac{\sin(2\tan^{-1} x)}{1+x^2} dx = \int \sin 2t \, dt = -\frac{1}{2} \cos 2t + c = -\frac{1}{2} \cos(2\tan^{-1} x) + c \)
In simple words: The differential \( \frac{dx}{1+x^2} \) is the differential of \( \tan^{-1} x \) - make this substitution to convert a trig-inverse problem into a standard trigonometric integral.

Exam Tip: Remember the standard differentials: \( \frac{dx}{1+x^2} = d(\tan^{-1} x) \) and \( \frac{dx}{\sqrt{1-x^2}} = d(\sin^{-1} x) \).

Question 17. Evaluate: \( \int \frac{\tan x \sec^2 x}{(1-\tan^2 x)} dx \)
Answer: Let \( t = 1 - \tan^2 x \), so \( dt = -2\tan x \sec^2 x \, dx \) or \( \tan x \sec^2 x \, dx = -\frac{1}{2} dt \). Substituting:
\( \int \frac{\tan x \sec^2 x}{1-\tan^2 x} dx = \int \frac{-\frac{1}{2} dt}{t} = -\frac{1}{2} \ln|t| + c = -\frac{1}{2} \ln|1-\tan^2 x| + c \)
In simple words: Observe that \( \tan x \sec^2 x \) is (up to a constant) the derivative of \( \tan^2 x \) - use substitution to convert this into a logarithmic integral.

Exam Tip: When you see a trig function times its own derivative (like \( \tan x \) times \( \sec^2 x \)), always substitute that function.

Question 18. Evaluate: \( \int \frac{(x^4+1)}{(x^2+1)} dx \)
Answer: Rewrite the numerator: \( x^4 + 1 = (x^4 - 1) + 2 = (x^2-1)(x^2+1) + 2 \). Thus:
\( \frac{x^4+1}{x^2+1} = (x^2-1) + \frac{2}{x^2+1} \)
\( \int \frac{x^4+1}{x^2+1} dx = \int \left( x^2 - 1 + \frac{2}{x^2+1} \right) dx = \frac{x^3}{3} - x + 2\tan^{-1} x + c \)
In simple words: Use polynomial long division or algebraic manipulation to break the improper fraction into a polynomial plus a simpler fraction that you can integrate term by term.

Exam Tip: For rational integrals where the numerator has degree equal to or greater than the denominator, always perform polynomial division first.

Question 19. Evaluate: \( \int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}} dx \)
Answer: Note that \( \sin x = \cos\left(\frac{\pi}{2} - x\right) \). Using half-angle identities:
\( \frac{1-\sin x}{1+\sin x} = \frac{\sin^2\left(\frac{\pi}{4} - \frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)} = \tan^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \)
Therefore \( \sqrt{\frac{1-\sin x}{1+\sin x}} = \left|\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right| \), and:
\( \int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}} dx = \int \left(\frac{\pi}{4} - \frac{x}{2}\right) dx = \frac{\pi}{4} x - \frac{x^2}{4} + c \)
In simple words: Simplify the expression inside the inverse tangent using half-angle identities until you recognize an inverse tangent of a known angle.

Exam Tip: Half-angle and double-angle identities are essential tools for simplifying complex trigonometric expressions before integration.

Question 20. Evaluate: \( \int \log(1+x^2) dx \)
Answer: Use integration by parts: \( \int u \, dv = uv - \int v \, du \), with \( u = \log(1+x^2) \) and \( dv = dx \). Then \( du = \frac{2x}{1+x^2} dx \) and \( v = x \). Thus:
\( \int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx = x\log(1+x^2) - 2\int \frac{x^2}{1+x^2} dx \)
Rewrite \( \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \). Then:
\( \int \frac{x^2}{1+x^2} dx = \int \left(1 - \frac{1}{1+x^2}\right) dx = x - \tan^{-1} x \)
\( \int \log(1+x^2) dx = x\log(1+x^2) - 2(x - \tan^{-1} x) + c = x\log(1+x^2) - 2x + 2\tan^{-1} x + c \)
In simple words: Use integration by parts to convert a logarithm integral into a rational function, then use polynomial division to finish the calculation.

Exam Tip: For integrals of logarithms, inverse trig functions, or products of trig and exponential functions, integration by parts is usually the first choice.

Question 21. Evaluate: \( \int \cos x \cos 3x \, dx \)
Answer: Apply the product-to-sum identity \( 2\cos A \cos B = \cos(A+B) + \cos(A-B) \):
\( 2\cos x \cos 3x = \cos 4x + \cos 2x \)
\( \int \cos x \cos 3x \, dx = \frac{1}{2} \int (\cos 4x + \cos 2x) dx = \frac{1}{2} \left( \frac{\sin 4x}{4} + \frac{\sin 2x}{2} \right) + c = \frac{\sin 4x}{8} + \frac{\sin 2x}{4} + c \)
In simple words: Convert a product of cosines into a sum of cosines using the product-to-sum formula, then integrate term by term.

Exam Tip: Master the product-to-sum and sum-to-product identities - they transform trigonometric products into integrals you can handle easily.

Question 22. Evaluate: \( \int \sin 3x \sin x \, dx \)
Answer: Use the product-to-sum identity \( 2\sin A \sin B = \cos(A-B) - \cos(A+B) \):
\( 2\sin 3x \sin x = \cos 2x - \cos 4x \)
\( \int \sin 3x \sin x \, dx = \frac{1}{2} \int (\cos 2x - \cos 4x) dx = \frac{1}{2} \left( \frac{\sin 2x}{2} - \frac{\sin 4x}{4} \right) + c = \frac{\sin 2x}{4} - \frac{\sin 4x}{8} + c \)
In simple words: Apply the product-to-sum rule to convert the sine product into a difference of cosines, then integrate each cosine separately.

Exam Tip: Keep the product-to-sum formulas on a reference card until they become automatic - they appear repeatedly in trig integrals.

Question 23. Evaluate: \( \int \tan^{-1}\sqrt{\frac{1-\cos(\pi/2-x)}{1+\cos(\pi/2-x)}} dx \)
Answer: Note that \( \cos\left(\frac{\pi}{2}-x\right) = \sin x \). Using half-angle identities:
\( \frac{1-\sin x}{1+\sin x} = \tan^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \)
Therefore \( \sqrt{\frac{1-\sin x}{1+\sin x}} = \left|\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right| \), and:
\( \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = \frac{\pi}{4} - \frac{x}{2} \)
\( \int \left(\frac{\pi}{4} - \frac{x}{2}\right) dx = \frac{\pi}{4} x - \frac{x^2}{4} + c \)
In simple words: Simplify the nested functions using angle identities, recognize the inverse tangent of an angle, then integrate the resulting linear expression.

Exam Tip: When inverse trig functions appear in integrals, look for ways to simplify the argument using standard identities - often the inverse cancels with a known angle.

Question 24. Evaluate: \( \int e^x(\tan x - \log \cos x) dx \)
Answer: Recognize that this integral fits the form \( \int e^x(f(x) + f'(x)) dx = e^x f(x) + c \). Here, if \( f(x) = -\log \cos x \), then:
\( f'(x) = \frac{\sin x}{\cos x} = \tan x \)
\( \int e^x(\tan x - \log \cos x) dx = -e^x \log \cos x + c \)
In simple words: When you see \( e^x \) times a sum where one term is the derivative of the other, use the special formula for exponential integrals.

Exam Tip: The formula \( \int e^x(f(x) + f'(x)) dx = e^x f(x) + c \) is a powerful shortcut - watch for it in exponential integrals.

Question 25. Evaluate: \( \int \frac{dx}{(1-\sin x)} \)
Answer: Multiply numerator and denominator by \( (1+\sin x) \):
\( \frac{1}{1-\sin x} = \frac{1+\sin x}{(1-\sin x)(1+\sin x)} = \frac{1+\sin x}{1-\sin^2 x} = \frac{1+\sin x}{\cos^2 x} = \sec^2 x + \sec x \tan x \)
\( \int \frac{dx}{1-\sin x} = \int (\sec^2 x + \sec x \tan x) dx = \tan x + \sec x + c \)
In simple words: Rationalize the denominator by multiplying by the conjugate, then express the result as a sum of standard trig integrals.

Exam Tip: Rationalizing denominators in trigonometric integrals often converts them into recognizable standard forms.

Question 26. Evaluate: \( \int x \cos x^2 dx \)
Answer: Let \( t = x^2 \), so \( dt = 2x \, dx \) or \( x \, dx = \frac{1}{2} dt \). Substituting:
\( \int x \cos x^2 dx = \int \cos t \cdot \frac{1}{2} dt = \frac{1}{2} \sin t + c = \frac{1}{2} \sin x^2 + c \)
In simple words: When the power of the variable and the coefficient of \( dx \) relate through a derivative, substitute the power itself.

Exam Tip: Recognize substitution pairs: if you see \( x^n \cos(x^{n+1}) \), substitute \( x^{n+1} \).

Question 27. Evaluate: \( \int \frac{xe^x}{(x+1)^2} dx \)
Answer: Rewrite the numerator as \( xe^x = e^x(x+1) - e^x \):
\( \frac{xe^x}{(x+1)^2} = \frac{e^x(x+1)}{(x+1)^2} - \frac{e^x}{(x+1)^2} = \frac{e^x}{x+1} - \frac{e^x}{(x+1)^2} \)
Using the formula \( \int e^x(f(x) + f'(x)) dx = e^x f(x) + c \) with \( f(x) = \frac{1}{x+1} \):
\( \int \frac{xe^x}{(x+1)^2} dx = \frac{e^x}{x+1} + c \)
In simple words: Rewrite the numerator strategically so that \( e^x \) multiplies a function and its derivative, then apply the exponential integration shortcut.

Exam Tip: When facing a ratio of polynomials times \( e^x \), look for ways to express it in the form \( e^x(f + f') \).

Question 28. Evaluate: \( \int \frac{\cot x}{\sqrt{\sin x}} dx \)
Answer: Rewrite as \( \frac{\cot x}{\sqrt{\sin x}} = \frac{\cos x}{(\sin x)^{3/2}} \). Let \( t = \sin x \), so \( dt = \cos x \, dx \). Substituting:
\( \int \frac{\cos x}{(\sin x)^{3/2}} dx = \int \frac{dt}{t^{3/2}} = \int t^{-3/2} dt = \frac{t^{-1/2}}{-1/2} + c = -\frac{2}{\sqrt{t}} + c = -\frac{2}{\sqrt{\sin x}} + c \)
In simple words: Convert radicals and quotients to exponential form, substitute the trig function, then use the power rule.

Exam Tip: Radicals and fractional powers can be tricky - always convert to exponential notation before integrating.

Question 29. Evaluate: \( \int \frac{\sec^2 x}{\cos \sec^2 x} dx \)
Answer: Note that \( \frac{\sec^2 x}{\cos \sec^2 x} = \tan^2 x \). Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\( \int \tan^2 x \, dx = \int (\sec^2 x - 1) dx = \tan x - x + c \)
In simple words: Simplify the expression first, then apply a standard identity to convert it into basic integrals.

Exam Tip: Always simplify and use identities before integrating - this often reduces a complex integral to a straightforward one.

Question 30. Evaluate: \( \int \frac{dx}{(\sqrt{x+2}+\sqrt{x+1})} \)
Answer: Rationalize by multiplying numerator and denominator by \( (\sqrt{x+2}-\sqrt{x+1}) \):
\( \frac{1}{\sqrt{x+2}+\sqrt{x+1}} = \frac{\sqrt{x+2}-\sqrt{x+1}}{(\sqrt{x+2})^2-(\sqrt{x+1})^2} = \frac{\sqrt{x+2}-\sqrt{x+1}}{1} \)
\( \int \left(\sqrt{x+2}-\sqrt{x+1}\right) dx = \frac{2}{3}(x+2)^{3/2} - \frac{2}{3}(x+1)^{3/2} + c \)
In simple words: Multiply by the conjugate to clear the radicals from the denominator, then integrate each radical separately using the power rule.

Exam Tip: Conjugate rationalization is essential when radicals appear in denominators - it often converts the problem into basic polynomial or root integrals.

Question 31. Evaluate: \( \int 2^x dx \)
Answer: Using the standard formula \( \int a^x dx = \frac{a^x}{\ln a} + c \) with \( a = 2 \):
\( \int 2^x dx = \frac{2^x}{\ln 2} + c \)
In simple words: For any exponential function with a constant base (other than \( e \)), divide by the natural logarithm of the base.

Exam Tip: The formula for exponential integrals is \( \int a^x dx = \frac{a^x}{\ln a} + c \) and \( \int e^x dx = e^x + c \) - do not confuse them.

Question 32. Evaluate: \( \int \frac{(1+\tan x)}{(x+\log \sec x)} dx \)
Answer: Let \( t = x + \log(\sec x) \), so \( dt = (1 + \tan x) dx \). Substituting:
\( \int \frac{1+\tan x}{x+\log \sec x} dx = \int \frac{dt}{t} = \ln|t| + c = \ln|x + \log(\sec x)| + c \)
In simple words: The numerator is the derivative of the expression in the denominator - recognize this pattern and integrate as a logarithm.

Exam Tip: Always compute the derivative of the denominator and check whether the numerator matches it - this is often the key to solving the problem instantly.

Question 33. Evaluate: \( \int \frac{\sec^2(\log x)}{x} dx \)
Answer: Let \( t = \log x \), so \( \frac{1}{x} dx = dt \). Substituting:
\( \int \frac{\sec^2(\log x)}{x} dx = \int \sec^2 t \, dt = \tan t + c = \tan(\log x) + c \)
In simple words: The differential \( \frac{dx}{x} \) is the differential of \( \log x \) - substitute the logarithm to turn this into a basic secant-squared integral.

Exam Tip: Memorize: \( \int \sec^2 u \, du = \tan u + c \) and \( \int \csc^2 u \, du = -\cot u + c \).

Question 34. Evaluate: \( \int (2x+1)\sqrt{x^2+x+1} dx \)
Answer: Let \( t = x^2 + x + 1 \), so \( dt = (2x+1) dx \). Substituting:
\( \int (2x+1)\sqrt{x^2+x+1} dx = \int \sqrt{t} \, dt = \int t^{1/2} dt = \frac{t^{3/2}}{3/2} + c = \frac{2}{3}(x^2+x+1)^{3/2} + c \)
In simple words: Recognize that the coefficient \( 2x+1 \) is the derivative of \( x^2+x+1 \) - substitute and apply the power rule.

Exam Tip: When you see a function raised to a power and its derivative nearby, that function is your substitution.

Question 35. Evaluate: \( \int \frac{dx}{\sqrt{9x^2+16}} \)
Answer: Recognize this as a form of \( \int \frac{dx}{\sqrt{(ax)^2+b^2}} = \frac{1}{a} \ln|ax + \sqrt{(ax)^2+b^2}| + c \). Here \( a = 3 \) and \( b = 4 \):
\( \int \frac{dx}{\sqrt{9x^2+16}} = \frac{1}{3} \ln|3x + \sqrt{9x^2+16}| + c \)
In simple words: Identify the form \( \sqrt{(ax)^2 + b^2} \) and apply the standard logarithmic integral formula.

Exam Tip: Memorize the standard forms for \( \sqrt{a^2 - b^2x^2} \), \( \sqrt{a^2 + b^2x^2} \), and \( \sqrt{b^2x^2 - a^2} \) - they appear frequently.

Question 36. Evaluate: \( \int \frac{dx}{\sqrt{4-9x^2}} \)
Answer: Recognize this as a form of \( \int \frac{dx}{\sqrt{b^2-(ax)^2}} = \frac{1}{a} \sin^{-1}\left(\frac{ax}{b}\right) + c \). Here \( a = 3 \) and \( b = 2 \):
\( \int \frac{dx}{\sqrt{4-9x^2}} = \frac{1}{3} \sin^{-1}\left(\frac{3x}{2}\right) + c \)
In simple words: Identify the form \( \sqrt{b^2 - (ax)^2} \) and apply the inverse sine integral formula.

Exam Tip: The inverse sine formula appears when the discriminant is negative - watch for the minus sign in the radical.

Question 37. Evaluate: \( \int \frac{dx}{\sqrt{4x^2-25}} \)
Answer: Recognize this as a form of \( \int \frac{dx}{\sqrt{(ax)^2-b^2}} = \frac{1}{a} \ln|ax + \sqrt{(ax)^2-b^2}| + c \). Here \( a = 2 \) and \( b = 5 \):
\( \int \frac{dx}{\sqrt{4x^2-25}} = \frac{1}{2} \ln|2x + \sqrt{4x^2-25}| + c \)
In simple words: Identify the form \( \sqrt{(ax)^2 - b^2} \) and apply the standard logarithmic integral formula.

Exam Tip: This formula applies when the coefficient of \( x^2 \) is larger than the constant - the result involves a logarithm, not an inverse trig function.

Question 38. Evaluate: \( \int \frac{\sin(2\tan^{-1} x)}{(1+x^2)} dx \)
Answer: Let \( t = \tan^{-1} x \), so \( \frac{1}{1+x^2} dx = dt \). Substituting:
\( \int \frac{\sin(2\tan^{-1} x)}{1+x^2} dx = \int \sin 2t \, dt = -\frac{1}{2} \cos 2t + c = -\frac{1}{2} \cos(2\tan^{-1} x) + c \)
In simple words: Substitute the inverse tangent to convert a trig-inverse problem into a standard trigonometric integral.

Exam Tip: The differential \( \frac{dx}{1+x^2} \) is the derivative of \( \tan^{-1} x \) - always use this when you see this combination.

Question 38. Evaluate: \( \int\sqrt{4 - x^2} \, dx \)
Answer: We apply the standard formula for integrals of the form \( \int\sqrt{a^2 - x^2} \, dx \). The result is:
\( \int\sqrt{4 - x^2} \, dx = \int\sqrt{2^2 - x^2} \, dx = \frac{x}{2}\sqrt{4 - x^2} + 2 \sin^{-1}\frac{x}{2} + c \)

Exam Tip: Remember the inverse sine formula: \( \int\sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + c \). Identify \( a = 2 \) and substitute accordingly.

Question 39. Evaluate: \( \int\sqrt{9 + x^2} \, dx \)
Answer: We use the standard formula for integrals of the form \( \int\sqrt{a^2 + x^2} \, dx \). The result is:
\( \int\sqrt{9 + x^2} \, dx = \int\sqrt{3^2 + x^2} \, dx = \frac{x}{2}\sqrt{9 + x^2} + \frac{9}{2}\log\left|x + \sqrt{9 + x^2}\right| + c \)

Exam Tip: For expressions of the form \( \sqrt{a^2 + x^2} \), the formula involves both a product term and a logarithmic term. Take care to include the absolute value in the logarithm.

Question 40. Evaluate: \( \int\sqrt{x^2 - 16} \, dx \)
Answer: We apply the formula for integrals of the form \( \int\sqrt{x^2 - a^2} \, dx \). Since \( 16 = 4^2 \), we get:
\( \int\sqrt{x^2 - 16} \, dx = \int\sqrt{x^2 - 4^2} \, dx = \frac{x}{2}\sqrt{x^2 - 16} - 8\log\left|x + \sqrt{x^2 - 16}\right| + c \)

Exam Tip: For \( \sqrt{x^2 - a^2} \), the formula carries a logarithmic term with a negative coefficient. Always ensure the argument of the logarithm remains positive.

Question 1. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{9 + x^2} = ? \)
(a) \( \tan^{-1}\frac{x}{3} + C \)
(b) \( \frac{1}{3}\tan^{-1}\frac{x}{3} + C \)
(c) \( 3\tan^{-1}\frac{x}{3} + C \)
(d) none of these
Answer: (b) \( \frac{1}{3}\tan^{-1}\frac{x}{3} + C \)
In simple words: Recognize that \( 9 + x^2 = 3^2 + x^2 \). Using the formula \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C \) with \( a = 3 \), we obtain the result by substituting.

Exam Tip: Always rewrite the denominator as a sum of squares. The coefficient multiplying the inverse tangent comes from \( \frac{1}{a} \), not from the expression itself.

Question 2. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{4 + 16x^2} = ? \)
(a) \( \frac{1}{32}\tan^{-1}4x + C \)
(b) \( \frac{1}{16}\tan^{-1}\frac{x}{2} + C \)
(c) \( \frac{1}{8}\tan^{-1}2x + C \)
(d) \( \frac{1}{4}\tan^{-1}\frac{x}{2} + C \)
Answer: (c) \( \frac{1}{8}\tan^{-1}2x + C \)
In simple words: Factor out 4 from the denominator to get \( 4(1 + 4x^2) = (2^2 + (2x)^2) \). Let \( t = 2x \), so \( dt = 2dx \). Then evaluate using the standard formula and substitute back.

Exam Tip: Factor constants from the denominator before applying the standard formula. Use substitution to handle coefficients of \( x \) inside the squared term.

Question 3. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{9 + 4x^2} = ? \)
(a) \( \frac{1}{6}\tan^{-1}\frac{2x}{3} + C \)
(b) \( \frac{1}{6}\tan^{-1}\frac{3x}{2} + C \)
(c) \( \frac{1}{4}\tan^{-1}\frac{3x}{2} + C \)
(d) none of these
Answer: (a) \( \frac{1}{6}\tan^{-1}\frac{2x}{3} + C \)
In simple words: Rewrite the denominator as \( 3^2 + (2x)^2 \). Let \( t = 2x \), giving \( dt = 2dx \). Apply the inverse tangent formula with \( a = 3 \) and substitute back to finish.

Exam Tip: When the denominator contains both a constant term and a term with \( x \), factor and rewrite as a sum of squares before substituting.

Question 4. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin x}{1 + \cos^2 x} dx = ? \)
(a) \( -\tan^{-1}(\cos x) + C \)
(b) \( \cot^{-1}(\cos x) + C \)
(c) \( -\cot^{-1}(\cos x) + C \)
(d) \( \tan^{-1}(\cos x) + C \)
Answer: (a) \( -\tan^{-1}(\cos x) + C \)
In simple words: Let \( t = \cos x \), so \( dt = -\sin x \, dx \). The integral becomes \( -\int \frac{dt}{1 + t^2} \), which evaluates to \( -\tan^{-1}t + C \). Substitute back to complete the solution.

Exam Tip: When you see a trigonometric function in the numerator, look for its derivative in relation to the denominator. The substitution will often simplify directly.

Question 5. Mark (√) against the correct answer in each of the following: \( \int \frac{\cos x}{1 + \sin^2 x} dx = ? \)
(a) \( -\tan^{-1}(\sin x) + C \)
(b) \( \tan^{-1}(\cos x) + C \)
(c) \( \tan^{-1}(\sin x) + C \)
(d) \( -\tan^{-1}(\cos x) + C \)
Answer: (c) \( \tan^{-1}(\sin x) + C \)
In simple words: Use the substitution \( t = \sin x \), which gives \( dt = \cos x \, dx \). The integral simplifies to \( \int \frac{dt}{1 + t^2} = \tan^{-1}t + C \). Replace \( t \) with \( \sin x \) to finish.

Exam Tip: Compare the numerator with the derivative of the argument in the denominator to identify the right substitution immediately.

Question 6. Mark (√) against the correct answer in each of the following: \( \int \frac{e^x}{e^{2x} + 1} dx = ? \)
(a) \( \cot^{-1}(e^x) + C \)
(b) \( \tan^{-1}(e^x) + C \)
(c) \( 2\tan^{-1}(e^x) + C \)
(d) none of these
Answer: (b) \( \tan^{-1}(e^x) + C \)
In simple words: Let \( t = e^x \), so \( dt = e^x \, dx \). The denominator becomes \( t^2 + 1 \). The integral transforms to \( \int \frac{dt}{t^2 + 1} = \tan^{-1}t + C \). Substitute back to get the final answer.

Exam Tip: Exponential functions in both the numerator and denominator often simplify through substitution. Always look for the differential relationship.

Question 7. Mark (√) against the correct answer in each of the following: \( \int \frac{3x^5}{1 + x^{12}} dx = ? \)
(a) \( \tan^{-1}x^6 + C \)
(b) \( \frac{1}{4}\tan^{-1}x^6 + C \)
(c) \( \frac{1}{2}\tan^{-1}x^6 + C \)
(d) none of these
Answer: (c) \( \frac{1}{2}\tan^{-1}x^6 + C \)
In simple words: Let \( t = x^6 \), so \( dt = 6x^5 \, dx \), which means \( 3x^5 \, dx = \frac{dt}{2} \). The integral becomes \( \frac{1}{2}\int \frac{dt}{1 + t^2} = \frac{1}{2}\tan^{-1}t + C \). Replace \( t \) with \( x^6 \) to obtain the result.

Exam Tip: When a power of \( x \) appears in the numerator, compare it with the exponent in the denominator to set up the substitution and adjust the coefficient.

Question 8. Mark (√) against the correct answer in each of the following: \( \int \frac{2x^3}{4 + x^8} dx = ? \)
(a) \( \frac{1}{2}\tan^{-1}\frac{x^4}{2} + C \)
(b) \( \frac{1}{4}\tan^{-1}\frac{x^4}{2} + C \)
(c) \( \frac{1}{2}\tan^{-1}x^4 + C \)
(d) none of these
Answer: (b) \( \frac{1}{4}\tan^{-1}\frac{x^4}{2} + C \)
In simple words: Set \( t = x^4 \), so \( dt = 4x^3 \, dx \), which gives \( 2x^3 \, dx = \frac{dt}{2} \). The denominator becomes \( 4 + t^2 = 2^2 + t^2 \). The integral becomes \( \frac{1}{2}\int \frac{dt}{4 + t^2} = \frac{1}{2} \cdot \frac{1}{2}\tan^{-1}\frac{t}{2} + C \). Substitute \( t = x^4 \) back.

Exam Tip: Pay attention to the relationship between the power in the numerator and the exponent in the denominator. Adjust constants carefully during substitution.

Question 9. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{x^2 + 4x + 8} = ? \)
(a) \( \frac{1}{2}\tan^{-1}\left(\frac{x + 1}{2}\right) + C \)
(b) \( \frac{1}{2}\tan^{-1}\left(\frac{x + 2}{2}\right) + C \)
(c) \( \frac{1}{2}\tan^{-1}(x + 2) + C \)
(d) \( \tan^{-1}\left(\frac{x + 2}{2}\right) + C \)
Answer: (b) \( \frac{1}{2}\tan^{-1}\left(\frac{x + 2}{2}\right) + C \)
In simple words: Complete the square in the denominator: \( x^2 + 4x + 8 = (x + 2)^2 + 4 \). Let \( t = x + 2 \), so \( dt = dx \). The integral transforms to \( \int \frac{dt}{t^2 + 2^2} = \frac{1}{2}\tan^{-1}\frac{t}{2} + C \). Substitute back to finish.

Exam Tip: Completing the square is essential when the denominator has both linear and quadratic terms. Express the result as a sum of a perfect square and a constant.

Question 10. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{2x^2 + x + 3} = ? \)
(a) \( \frac{1}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
(b) \( -\frac{1}{\sqrt{23}}\tan^{-1}\left(\frac{x + 1}{\sqrt{23}}\right) + C \)
(c) \( \frac{2}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
(d) none of these
Answer: (a) \( \frac{1}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
In simple words: Complete the square: \( 2x^2 + x + 3 = 2\left[\left(x + \frac{1}{4}\right)^2 + \frac{23}{16}\right] \). Let \( t = x + \frac{1}{4} \), so \( dt = dx \). The integral becomes \( \frac{1}{2}\int \frac{dt}{t^2 + \frac{23}{16}} \). Apply the inverse tangent formula and substitute back.

Exam Tip: When the coefficient of \( x^2 \) is not 1, factor it out completely before completing the square. This prevents errors in identifying \( a \) in the formula.

Question 11. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{2x^2 + x + 3} = ? \)
(a) \( \frac{2}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
(b) \( \frac{1}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
(c) \( \frac{4}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
(d) none of these
Answer: (b) \( \frac{1}{\sqrt{23}}\tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \)
In simple words: Completing the square yields \( 2x^2 + x + 3 = 2\left[\left(x + \frac{1}{4}\right)^2 + \frac{23}{16}\right] \). Set \( t = x + \frac{1}{4} \), so \( dt = dx \). Transform the integral to \( \frac{1}{2}\int \frac{dt}{t^2 + \left(\frac{\sqrt{23}}{4}\right)^2} \) and use the standard formula.

Exam Tip: After completing the square, the constant term under the square root in the denominator determines the coefficient in the final inverse tangent formula.

Question 12. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{e^x + e^{-x}} = ? \)
(a) \( \tan^{-1}(e^x) + C \)
(b) \( \tan^{-1}(e^{-x}) + C \)
(c) \( -\tan^{-1}(e^{-x}) + C \)
(d) none of these
Answer: (a) \( \tan^{-1}(e^x) + C \)
In simple words: Let \( t = e^x \), so \( dt = e^x \, dx \). Rewrite the denominator as \( e^x + e^{-x} = e^x + \frac{1}{e^x} = \frac{(e^x)^2 + 1}{e^x} = \frac{t^2 + 1}{t} \). The integral becomes \( \int \frac{t}{t^2 + 1} \cdot \frac{dt}{t} = \int \frac{dt}{t^2 + 1} = \tan^{-1}t + C \). Replace \( t \) with \( e^x \).

Exam Tip: When hyperbolic expressions appear, use substitution to convert them into standard forms. Simplify the denominator carefully after substitution.

Question 13. Mark (√) against the correct answer in each of the following: \( \int \frac{x^2 - 1}{x^2 + 4} dx = ? \)
(a) \( x - 5\tan^{-1}\frac{x}{2} + C \)
(b) \( x - \frac{5}{2}\tan^{-1}\frac{x}{2} + C \)
(c) \( x - \frac{5}{2}\tan^{-1}\frac{5x}{2} + C \)
(d) none of these
Answer: (b) \( x - \frac{5}{2}\tan^{-1}\frac{x}{2} + C \)
In simple words: Divide to rewrite: \( \frac{x^2 - 1}{x^2 + 4} = 1 - \frac{5}{x^2 + 4} \). Split the integral: \( \int 1 \, dx - 5\int \frac{dx}{x^2 + 4} \). The first part gives \( x \). For the second part, use the inverse tangent formula with \( a = 2 \) to get \( -5 \cdot \frac{1}{2}\tan^{-1}\frac{x}{2} \).

Exam Tip: Decompose rational functions by polynomial division when the numerator degree equals or exceeds the denominator degree. This separates simpler integrals.

Question 14. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{4 + 9x^2} = ? \)
(a) \( \frac{2}{3}\tan^{-1}\frac{3x}{2} + C \)
(b) \( \frac{1}{6}\tan^{-1}3x + C \)
(c) \( \frac{1}{6}\tan^{-1}\frac{3x}{2} + C \)
(d) none of these
Answer: (c) \( \frac{1}{6}\tan^{-1}\frac{3x}{2} + C \)
In simple words: Rewrite the denominator as \( 4 + 9x^2 = 2^2 + (3x)^2 \). Let \( t = 3x \), giving \( dt = 3dx \), so \( dx = \frac{dt}{3} \). The integral becomes \( \frac{1}{3}\int \frac{dt}{4 + t^2} = \frac{1}{3} \cdot \frac{1}{2}\tan^{-1}\frac{t}{2} + C \). Substitute \( t = 3x \) back and simplify.

Exam Tip: Factor the denominator to identify both the constant and the coefficient of \( x \) separately. This makes substitution and formula application clearer.

Question 15. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{4x^2 - 4x + 3} = ? \)
(a) \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{2}}\right) + C \)
(b) \( \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{2}}\right) + C \)
(c) \( -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{2}}\right) + C \)
(d) none of these
Answer: (b) \( \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{2x - 1}{\sqrt{2}}\right) + C \)
In simple words: Complete the square: \( 4x^2 - 4x + 3 = 4\left(x^2 - x + \frac{3}{4}\right) = 4\left[\left(x - \frac{1}{2}\right)^2 + \frac{1}{2}\right] \). Let \( t = x - \frac{1}{2} \), so \( dt = dx \). The integral becomes \( \frac{1}{4}\int \frac{dt}{t^2 + \frac{1}{2}} \). Apply the inverse tangent formula with \( a = \frac{1}{\sqrt{2}} \) and substitute back.

Exam Tip: When completing the square with a coefficient on \( x^2 \), factor it out first to ensure accurate calculations of the linear and constant terms.

Question 16. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sin^4 x + \cos^4 x} = ? \)
(a) \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^2 x - 1}{\sqrt{2}\tan x}\right) + C \)
(b) \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^2 x - 1}{\tan x}\right) + C \)
(c) \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1}{\sqrt{2}\tan x}\right) + C \)
(d) None of these
Answer: (a) \( \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^2 x - 1}{\sqrt{2}\tan x}\right) + C \)
In simple words: Rewrite the denominator as \( \sin^4 x + \cos^4 x = \cos^4 x(\tan^4 x + 1) \). Divide numerator and denominator by \( \cos^4 x \) to get \( \sec^4 x(\tan^4 x + 1) \). Let \( t = \tan x \), so \( dt = \sec^2 x \, dx \). The integral transforms to \( \int \frac{\sec^2 x \, sec^2 x}{t^4 + 1} dx \), which simplifies using the substitution and yields the inverse tangent result.

Exam Tip: For integrals with fourth powers of trigonometric functions, divide by the highest power of cosine to convert to tangent and secant. This converts the problem into a polynomial-based integral.

Question 17. Mark (√) against the correct answer in each of the following: \( \int \frac{(x^2 + 1)}{(x^4 + x^2 + 1)} dx = ? \)
(a) \( \tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}}\right) + C \)
(b) \( \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}}\right) + C \)
(c) \( \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}x}\right) + C \)
(d) none of these
Answer: (b) \( \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}}\right) + C \)
In simple words: Divide numerator and denominator by \( x^2 \) to rewrite the integrand as \( \frac{1 + x^{-2}}{x^2 + 1 + x^{-2}} \). Let \( t = x - x^{-1} \), so \( dt = (1 + x^{-2}) dx \). The denominator becomes \( (x - x^{-1})^2 + 3 = t^2 + 3 \). The integral simplifies to \( \int \frac{dt}{t^2 + 3} = \frac{1}{\sqrt{3}}\tan^{-1}\frac{t}{\sqrt{3}} + C \). Substitute back to complete.

Exam Tip: When dealing with expressions involving both \( x \) and its reciprocal, divide through by \( x^2 \) and look for a substitution of the form \( x - \frac{1}{x} \) or \( x + \frac{1}{x} \).

Question 18. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx = ? \)
(a) \( \tan^{-1}(\tan^2 x) + C \)
(b) \( x^2 + C \)
(c) \( -\tan^{-1}(\tan^2 x) + C \)
(d) none of these
Answer: (a) \( \tan^{-1}(\tan^2 x) + C \)
In simple words: Rewrite \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x = 1 - \frac{1}{2}(1 - \cos^2 2x)/2 = (\sin^2 2x + 1)^2 - 1 \) after simplification. Let \( t = \sec^2 x - 1 = \tan^2 x \), so \( dt = 2\sec x \sec x \tan x \, dx \). After algebra, the integral evaluates to \( \tan^{-1}t + C = \tan^{-1}(\tan^2 x) + C \).

Exam Tip: When \( \sin 2x \) appears in the numerator with powers of sine and cosine in the denominator, consider substituting a related trigonometric function. The double angle formula often simplifies the calculation.

Question 20. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(1-9x^{2}\right)} = ? \)
(a) \( \frac{1}{3}\log\left|\frac{1+3x}{1-3x}\right| + C \)
(b) \( \frac{1}{3}\log\left|\frac{1-3x}{1+3x}\right| + C \)
(c) \( \frac{1}{6}\log\left|\frac{1+3x}{1-3x}\right| + C \)
(d) \( \frac{1}{6}\log\left|\frac{1-3x}{1+3x}\right| + C \)
Answer: (b) \( \frac{1}{3}\log\left|\frac{1-3x}{1+3x}\right| + C \)
In simple words: Set \( 3x = t \), so \( 3dx = dt \). This simplifies the integral into the form \( \frac{1}{a^2-x^2} \), which yields a logarithmic result with the given coefficients and absolute value ratio.

Exam Tip: Always identify when the denominator fits the \( a^2 - (bx)^2 \) pattern and substitute accordingly to match standard integral formulas.

Question 21. Mark (√) against the correct answer in each of the following: \( \int \frac{x^{2}}{\left(1-x^{6}\right)} dx = ? \)
(a) \( \frac{1}{6}\log\left|\frac{1+x^3}{1-x^3}\right| + C \)
(b) \( \frac{1}{6}\log\left|\frac{1-x^3}{1+x^3}\right| + C \)
(c) \( \frac{1}{3}\log\left|\frac{1-x^3}{1+x^3}\right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{6}\log\left|\frac{1-x^3}{1+x^3}\right| + C \)
In simple words: Use the substitution \( x^3 = t \), which gives \( 3x^2 dx = dt \), so \( x^2 dx = \frac{dt}{3} \). This transforms the integral into the standard form \( \frac{1}{1^2 - t^2} \), leading to a logarithm with the correct coefficient.

Exam Tip: Recognize that \( x^6 = (x^3)^2 \) and apply substitution on the higher power; this reduces the integral to a familiar pattern.

Question 22. Mark (√) against the correct answer in each of the following: \( \int \frac{x}{\left(1-x^{4}\right)} dx = ? \)
(a) \( \frac{1}{4}\log\left|\frac{1+x^2}{1-x^2}\right| + C \)
(b) \( \frac{1}{4}\log\left|\frac{1-x^2}{1+x^2}\right| + C \)
(c) \( \frac{1}{2}\log\left|\frac{1+x^2}{1-x^2}\right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{4}\log\left|\frac{1+x^2}{1-x^2}\right| + C \)
In simple words: Let \( x^2 = t \), so \( 2x dx = dt \), giving \( x dx = \frac{dt}{2} \). Since \( 1 - x^4 = 1 - (x^2)^2 = 1 - t^2 \), the integral reduces to \( \frac{1}{2} \int \frac{dt}{1-t^2} \), which evaluates to the logarithmic form shown.

Exam Tip: When you see \( x^4 \) in the denominator and \( x \) in the numerator, always consider substituting the even power of \( x \) to simplify.

Question 23. Mark (√) against the correct answer in each of the following: \( \int \frac{x^{2}}{\left(a^{6}-x^{6}\right)} dx = ? \)
(a) \( \frac{1}{3a^3}\log\left|\frac{a^3+x^3}{a^3-x^3}\right| + C \)
(b) \( \frac{1}{6a^3}\log\left|\frac{a^3+x^3}{a^3-x^3}\right| + C \)
(c) \( \frac{1}{6a^3}\log\left|\frac{a^3-x^3}{a^3+x^3}\right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{6a^3}\log\left|\frac{a^3+x^3}{a^3-x^3}\right| + C \)
In simple words: Use \( x^3 = t \), so \( 3x^2 dx = dt \), giving \( x^2 dx = \frac{dt}{3} \). The integral becomes \( \frac{1}{3} \int \frac{dt}{(a^3)^2 - t^2} \), which applies the standard \( a^2 - x^2 \) formula to produce the final logarithmic result.

Exam Tip: Identify when both the numerator exponent and denominator base powers are divisible by 3; this signals a \( x^3 \) substitution.

Question 24. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(3-2x-x^{2}\right)} = ? \)
(a) \( \frac{1}{4}\log\left|\frac{3+x}{3-x}\right| + C \)
(b) \( \frac{1}{4}\log\left|\frac{1+x}{1-x}\right| + C \)
(c) \( \frac{1}{4}\log\left|\frac{3+x}{1-x}\right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{4}\log\left|\frac{3+x}{1-x}\right| + C \)
In simple words: Complete the square in the denominator: \( 3 - 2x - x^2 = 4 - (x+1)^2 \). Substitute \( x + 1 = t \) to obtain \( \int \frac{dt}{2^2 - t^2} \). Apply the standard formula for differences of squares to get the logarithmic form.

Exam Tip: When facing a quadratic expression in the denominator with a mix of positive and negative terms, always complete the square first to reveal the standard form.

Question 25. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(\cos^{2}x - 3\sin^{2}x\right)} = ? \)
(a) \( \frac{1}{\sqrt{3}}\log\left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right| + C \)
(b) \( \frac{1}{\sqrt{3}}\log\left|\frac{1-\sqrt{3}\tan x}{1+\sqrt{3}\tan x}\right| + C \)
(c) \( \frac{1}{2\sqrt{3}}\log\left|\frac{1+\sqrt{3}\tan x}{1-\sqrt{3}\tan x}\right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{2\sqrt{3}}\log\left|\frac{1+\sqrt{3}\tan x}{1-\sqrt{3}\tan x}\right| + C \)
In simple words: Factor out \( \cos^2 x \) from the denominator to get \( \cos^2 x(1 - 3\tan^2 x) \). Divide numerator and denominator by \( \cos^2 x \) to obtain \( \sec^2 x \) in the numerator. Using \( \sqrt{3}\tan x = t \), the integral matches the standard \( 1 - t^2 \) formula.

Exam Tip: For integrals with mixed trigonometric functions in the denominator, always factor out a power of cosine and convert to a tangent substitution.

Question 26. Mark (√) against the correct answer in each of the following: \( \int \frac{\cosec^{2}x}{\left(1-\cot^{2}x\right)} dx = ? \)
(a) \( \frac{1}{2}\log\left|\frac{1+\cot x}{1-\cot x}\right| + C \)
(b) \( -\frac{1}{2}\log\left|\frac{1+\cot x}{1-\cot x}\right| + C \)
(c) \( \frac{1}{2}\log\left|\frac{1-\cot x}{1+\cot x}\right| + C \)
(d) None of the options
Answer: (b) \( -\frac{1}{2}\log\left|\frac{1+\cot x}{1-\cot x}\right| + C \)
In simple words: Set \( \cot x = t \), so \( -\cosec^2 x \, dx = dt \), giving \( \cosec^2 x \, dx = -dt \). The integral becomes \( \int \frac{-dt}{1-t^2} \), which yields a negative logarithmic result after applying the standard difference-of-squares formula.

Exam Tip: Pay careful attention to the negative sign that arises from differentiating cotangent; it affects the final answer's sign.

Question 27. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(4x^{2}-1\right)} = ? \)
(a) \( \frac{1}{2}\log\left|\frac{2x-1}{2x+1}\right| + C \)
(b) \( \frac{1}{2}\log\left|\frac{2x+1}{2x-1}\right| + C \)
(c) \( \frac{1}{4}\log\left|\frac{2x-1}{2x+1}\right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{4}\log\left|\frac{2x-1}{2x+1}\right| + C \)
In simple words: Recognize that \( 4x^2 - 1 = (2x)^2 - 1^2 \). Substitute \( 2x = t \) to get \( 2dx = dt \), so \( dx = \frac{dt}{2} \). The integral reduces to \( \frac{1}{2} \int \frac{dt}{t^2 - 1} \), which evaluates using the standard difference-of-squares formula.

Exam Tip: Always factor expressions like \( 4x^2 - 1 \) as a difference of squares to identify the appropriate substitution.

Question 28. Mark (√) against the correct answer in each of the following: \( \int \frac{x}{\left(x^{4}-16\right)} dx = ? \)
(a) \( \frac{1}{4}\log\left|\frac{x^2+4}{x^2-4}\right| + C \)
(b) \( \frac{1}{16}\log\left|\frac{x^2+4}{x^2-4}\right| + C \)
(c) \( \frac{1}{16}\log\left|\frac{x^2-4}{x^2+4}\right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{16}\log\left|\frac{x^2-4}{x^2+4}\right| + C \)
In simple words: Use \( x^2 = t \), so \( 2x \, dx = dt \), giving \( x \, dx = \frac{dt}{2} \). The integral becomes \( \frac{1}{2} \int \frac{dt}{t^2 - 16} = \frac{1}{2} \int \frac{dt}{t^2 - 4^2} \). Applying the standard formula yields the logarithmic result.

Exam Tip: When the numerator is \( x \) and the denominator contains \( x^4 \), substitute the quadratic term to reduce complexity.

Question 29. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(\sin^{2}x - 4\cos^{2}x\right)} = ? \)
(a) \( \frac{1}{4}\log\left|\frac{\tan x - 2}{\tan x + 2}\right| + C \)
(b) \( \frac{1}{4}\log\left|\frac{\tan x + 2}{\tan x - 2}\right| + C \)
(c) \( \frac{1}{4}\log\left|\frac{1 - \tan x}{1 + \tan x}\right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{4}\log\left|\frac{\tan x + 2}{\tan x - 2}\right| + C \)
In simple words: Factor out \( \cos^2 x \) from the denominator to obtain \( \cos^2 x(\tan^2 x - 4) \). Divide numerator and denominator by \( \cos^2 x \) to get \( \sec^2 x \). Set \( \tan x = t \) so \( \sec^2 x \, dx = dt \), and apply the difference-of-squares formula to the resulting integral.

Exam Tip: For trigonometric integrals with both sine and cosine squared, always factor and convert to tangent and secant.

Question 30. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\left(4\sin^{2}x + 5\cos^{2}x\right)} = ? \)
(a) \( \frac{1}{2}\tan^{-1}\left(\frac{\tan x}{\sqrt{5}}\right) + C \)
(b) \( \frac{1}{\sqrt{5}}\tan^{-1}\left(\frac{\tan x}{\sqrt{5}}\right) + C \)
(c) \( \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right) + C \)
(d) None of the options
Answer: (c) \( \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right) + C \)
In simple words: Factor \( \cos^2 x \) from the denominator to get \( \cos^2 x(4\tan^2 x + 5) \). Divide by \( \cos^2 x \) to obtain \( \sec^2 x \) in the numerator. Substituting \( 2\tan x = t \) yields \( 2\sec^2 x \, dx = dt \), and the integral becomes \( \frac{1}{2} \int \frac{dt}{t^2 + 5} \), which matches the arctangent formula.

Exam Tip: When coefficients of \( \sin^2 x \) and \( \cos^2 x \) differ and are both positive, expect an inverse tangent result.

Question 31. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin x}{\sin 3x} dx = ? \)
(a) \( \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\sin x}{\sqrt{3}-\sin x}\right| + C \)
(b) \( \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\cos x}{\sqrt{3}-\cos x}\right| + C \)
(c) \( \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right| + C \)
In simple words: Expand \( \sin 3x = 3\sin x - 4\sin^3 x = \sin x(3 - 4\sin^2 x) \). Divide numerator and denominator by \( \cos^2 x \) to convert to tangent form. Set \( \tan x = t \) and obtain an integral of the form \( \int \frac{dt}{3 - t^2} \), which evaluates to the logarithmic result shown.

Exam Tip: Use trigonometric identities to expand \( \sin 3x \) or \( \cos 3x \) in terms of lower powers before integrating.

Question 32. Mark (√) against the correct answer in each of the following: \( \int \frac{(x^{2}+1)}{(x^{4}+1)} dx = ? \)
(a) \( \frac{1}{2}\tan^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right) + C \)
(b) \( \frac{1}{2}\tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) + C \)
(c) \( \frac{1}{\sqrt{2}}\log\left|\frac{x^2+1}{x^2-1}\right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{2}\tan^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right) + C \)
In simple words: Divide numerator and denominator by \( x^2 \) to get \( \int \frac{1 + x^{-2}}{x^2 + x^{-2}} dx \). Substitute \( x - x^{-1} = t \), so \( (1 + x^{-2}) dx = dt \). This transforms the denominator into \( t^2 + 2 \), yielding an arctangent integral.

Exam Tip: For fourth-degree polynomials in the denominator with symmetric forms, divide by \( x^2 \) and use a substitution involving reciprocal terms.

Objective Questions II

Question 1. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{4-9x^{2}}} = ? \)
(a) \( \frac{1}{3}\sin^{-1}\frac{x}{3} + C \)
(b) \( \frac{2}{3}\sin^{-1}\left(\frac{2x}{3}\right) + C \)
(c) \( \frac{1}{3}\sin^{-1}\left(\frac{3x}{2}\right) + C \)
(d) None of the options
Answer: (b) \( \frac{2}{3}\sin^{-1}\left(\frac{2x}{3}\right) + C \)
In simple words: Rewrite the integral as \( \int \frac{1}{3} \frac{dx}{\sqrt{\frac{4}{9} - x^2}} = \int \frac{1}{3} \frac{dx}{\sqrt{\left(\frac{2}{3}\right)^2 - x^2}} \). Apply the standard inverse sine formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \).

Exam Tip: Always extract constants from under the square root and match the form to the standard inverse sine integral formula.

Question 2. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{16-4x^{2}}} = ? \)
(a) \( \frac{1}{2}\sin^{-1}\frac{x}{2} + C \)
(b) \( \frac{1}{4}\sin^{-1}\frac{x}{2} + C \)
(c) \( \frac{1}{2}\sin^{-1}\frac{x}{4} + C \)
(d) None of the options
Answer: (a) \( \frac{1}{2}\sin^{-1}\frac{x}{2} + C \)
In simple words: Factor to obtain \( \int \frac{1}{2} \frac{dx}{\sqrt{(2)^2 - x^2}} \). Using the standard inverse sine formula with \( a = 2 \), the result is \( \frac{1}{2}\sin^{-1}\left(\frac{x}{2}\right) + C \).

Exam Tip: Pull out coefficient factors from inside the square root before applying the inverse sine formula.

Question 3. Mark (√) against the correct answer in each of the following: \( \int \frac{\cos x}{\sqrt{4-\sin^{2}x}} dx = ? \)
(a) \( \sin^{-1}\frac{x}{2} + C \)
(b) \( \sin^{-1}\left(\frac{1}{2}\cos x\right) + C \)
(c) \( \sin^{-1}(2\sin x) + C \)
(d) \( \sin^{-1}\left(\frac{1}{2}\sin x\right) + C \)
Answer: (d) \( \sin^{-1}\left(\frac{1}{2}\sin x\right) + C \)
In simple words: Set \( \sin x = t \), so \( \cos x \, dx = dt \). The integral becomes \( \int \frac{dt}{\sqrt{4 - t^2}} = \int \frac{dt}{\sqrt{2^2 - t^2}} \), which equals \( \sin^{-1}\left(\frac{t}{2}\right) + C = \sin^{-1}\left(\frac{\sin x}{2}\right) + C \).

Exam Tip: When you see a trigonometric function in the numerator that matches the derivative of the argument in the denominator, always use that as your substitution.

Question 4. Mark (√) against the correct answer in each of the following: \( \int \frac{2^{x}}{\sqrt{1-4^{x}}} dx = ? \)
(a) \( \sin^{-1}(2^x) \log 2 + C \)
(b) \( \frac{\sin^{-1}\left(2^x\right)}{\log 2} + C \)
(c) \( \sin^{-1}(2^x) + C \)
(d) None of the options
Answer: (b) \( \frac{\sin^{-1}\left(2^x\right)}{\log 2} + C \)
In simple words: Set \( 2^x = t \), so \( 2^x \log 2 \, dx = dt \), giving \( 2^x dx = \frac{dt}{\log 2} \). Since \( 4^x = (2^2)^x = (2^x)^2 = t^2 \), the integral becomes \( \int \frac{1}{\log 2} \frac{dt}{\sqrt{1 - t^2}} = \frac{1}{\log 2} \sin^{-1}(t) + C \).

Exam Tip: Watch for exponential functions where the base of one term is the square of another (e.g., \( 4^x = (2^x)^2 \)); this structure calls for an exponential substitution.

Question 5. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{2x-x^{2}}} = ? \)
(a) \( \sin^{-1}(x + 1) + C \)
(b) \( \sin^{-1}(x - 2) + C \)
(c) \( \sin^{-1}(x - 1) + C \)
(d) None of the options
Answer: (c) \( \sin^{-1}(x - 1) + C \)
In simple words: Complete the square: \( 2x - x^2 = 1 - (x - 1)^2 \). The integral becomes \( \int \frac{dx}{\sqrt{1 - (x-1)^2}} = \sin^{-1}(x - 1) + C \) using the standard formula.

Exam Tip: Always complete the square when faced with a quadratic under a square root; this reveals the standard inverse sine form.

Question 6. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x(1-2x)}} = ? \)
(a) \( \frac{1}{\sqrt{2}}\sin^{-1}(2x - 1) + C \)
(b) \( \frac{1}{\sqrt{2}}\sin^{-1}(2x + 1) + C \)
(c) \( \frac{1}{\sqrt{2}}\sin^{-1}(4x + 1) + C \)
(d) \( \frac{1}{\sqrt{2}}\sin^{-1}(4x - 1) + C \)
Answer: (d) \( \frac{1}{\sqrt{2}}\sin^{-1}(4x - 1) + C \)
In simple words: Expand and complete the square: \( x(1 - 2x) = \frac{1}{8} - \left(x - \frac{1}{4}\right)^2 \). Through algebraic manipulation and substitution \( 4x - 1 = t \), the integral transforms into \( \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{\frac{1}{2} - \frac{t^2}{4}}} \), which simplifies to the inverse sine form shown.

Exam Tip: For expressions like \( x(1 - 2x) \), expand first, then complete the square carefully before choosing your substitution.

Question 7. Mark (√) against the correct answer in each of the following: \( \int \frac{3x^2}{\sqrt{9 - 16x^6}} dx = ? \)
(a) \( \frac{1}{4} \sin^{-1} \left( \frac{x^3}{3} \right) + C \)
(b) \( \frac{1}{4} \sin^{-1} \left( \frac{4x^3}{3} \right) + C \)
(c) \( 4 \sin^{-1} \left( \frac{x^3}{4} \right) + C \)
(d) None of the options
Answer: (b) \( \frac{1}{4} \sin^{-1} \left( \frac{4x^3}{3} \right) + C \)
In simple words: To solve this integral, let \( x^3 = t \), so \( 3x^2 dx = dt \). The integral becomes \( \frac{1}{3} \int \frac{dt}{\sqrt{9 - 16t^2}} \), which simplifies to \( \frac{1}{4} \sin^{-1} \left( \frac{4t}{3} \right) + c \). Substituting \( t = x^3 \) gives the final answer.

Exam Tip: Always recognize the pattern \( \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(u/a) + C \) and use substitution to match it.

Question 8. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{2 + 2x - x^2}} = ? \)
(a) \( \sin^{-1} \left( \frac{x - 1}{\sqrt{3}} \right) + C \)
(b) \( \sin^{-1} \left( \frac{x - 1}{\sqrt{2}} \right) + C \)
(c) \( \sin^{-1} \left( \sqrt{3}(x - 1) \right) + C \)
(d) None of the options
Answer: (b) \( \sin^{-1} \left( \frac{x - 1}{\sqrt{2}} \right) + C \)
In simple words: Complete the square in the denominator: \( 2 + 2x - x^2 = 2 - (x^2 - 2x) = 2 - (x - 1)^2 + 1 = 3 - (x - 1)^2 \). Rewrite as \( \sqrt{(\sqrt{2})^2 - (x - 1)^2} \) and apply the inverse sine formula.

Exam Tip: Completing the square is essential for integrals involving quadratic expressions under a square root.

Question 9. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{16 - 6x - x^2}} = ? \)
(a) \( \sin^{-1} \left( \frac{x - 3}{5} \right) + C \)
(b) \( \sin^{-1} \left( \frac{x + 3}{5} \right) + C \)
(c) \( \frac{1}{5} \sin^{-1} \left( x + 3 \right) + C \)
(d) None of the options
Answer: (b) \( \sin^{-1} \left( \frac{x + 3}{5} \right) + C \)
In simple words: Rearrange: \( 16 - 6x - x^2 = 16 - (x^2 + 6x) \). Complete the square: \( x^2 + 6x + 9 - 9 = (x + 3)^2 \), so \( 16 - 6x - x^2 = 25 - (x + 3)^2 \). Apply the inverse sine formula with numerator \( x + 3 \) and denominator 5.

Exam Tip: When the coefficient of \( x^2 \) is negative, rearrange so that you complete the square for the positive terms first.

Question 10. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x - x^2}} = ? \)
(a) \( \sin^{-1} \left( x - 1 \right) + C \)
(b) \( \sin^{-1} \left( x + 1 \right) + C \)
(c) \( \sin^{-1} \left( 2x - 1 \right) + C \)
(d) None of the options
Answer: (c) \( \sin^{-1} \left( 2x - 1 \right) + C \)
In simple words: Factor and complete the square: \( x - x^2 = x(1 - x) = \frac{1}{4} - (x - \frac{1}{2})^2 \). The integral becomes \( \int \frac{dx}{\sqrt{\frac{1}{4} - (x - \frac{1}{2})^2}} \), which gives \( \sin^{-1}(2x - 1) + C \).

Exam Tip: Always express the denominator as a perfect square minus another squared term to use the inverse sine formula.

Question 11. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{1 + 2x - 3x^2}} = ? \)
(a) \( \frac{1}{\sqrt{3}} \sin^{-1} \left( \frac{3x - 1}{2} \right) + C \)
(b) \( \frac{1}{\sqrt{2}} \sin^{-1} \left( \frac{2x - 1}{3} \right) + C \)
(c) \( \frac{1}{\sqrt{3}} \sin^{-1} \left( \frac{2x - 1}{3} \right) + C \)
(d) None of the options
Answer: (c) \( \frac{1}{\sqrt{3}} \sin^{-1} \left( \frac{2x - 1}{3} \right) + C \)
In simple words: Factor out the coefficient of \( x^2 \): \( 1 + 2x - 3x^2 = -3(x^2 - \frac{2}{3}x - \frac{1}{3}) \). Complete the square inside and simplify to get \( \frac{4}{3} - (x - \frac{1}{3})^2 \). After applying the formula and simplifying, the answer involves \( \frac{1}{\sqrt{3}} \) as the leading coefficient.

Exam Tip: When there is a negative coefficient on \( x^2 \), factor it out first, then complete the square and apply the inverse sine formula.

Question 12. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x^2 - 16}} = ? \)
(a) \( \sin^{-1} \left( \frac{x}{4} \right) + C \)
(b) \( \log \left| x + \sqrt{x^2 - 16} \right| + C \)
(c) \( \log \left| x - \sqrt{x^2 - 16} \right| + C \)
(d) None of the options
Answer: (b) \( \log \left| x + \sqrt{x^2 - 16} \right| + C \)
In simple words: This integral matches the standard form \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C \) with \( a = 4 \). Simply apply the formula directly.

Exam Tip: Memorize the standard forms for integrals of the type \( \int \frac{dx}{\sqrt{x^2 - a^2}} \) and \( \int \frac{dx}{\sqrt{a^2 - x^2}} \) to save time.

Question 13. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{4x^2 - 9}} = ? \)
(a) \( \frac{1}{2} \log \left| 2x + \sqrt{4x^2 - 9} \right| + C \)
(b) \( \frac{1}{4} \log \left| x + \sqrt{4x^2 - 9} \right| + C \)
(c) \( \log \left| 2x + \sqrt{4x^2 - 9} \right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{2} \log \left| 2x + \sqrt{4x^2 - 9} \right| + C \)
In simple words: Rewrite \( 4x^2 - 9 = (2x)^2 - 3^2 \). Let \( t = 2x \), so \( dt = 2 dx \) and \( dx = \frac{dt}{2} \). The integral becomes \( \frac{1}{2} \int \frac{dt}{\sqrt{t^2 - 9}} = \frac{1}{2} \log \left| t + \sqrt{t^2 - 9} \right| + C \). Substitute back \( t = 2x \).

Exam Tip: When the expression is a perfect square, use substitution to transform it into a standard form before applying the formula.

Question 14. Mark (√) against the correct answer in each of the following: \( \int \frac{x^2}{x^6 - 1} dx = ? \)
(a) \( \frac{1}{2} \log \left| x^3 + \sqrt{x^6 - 1} \right| + C \)
(b) \( \frac{1}{3} \log \left| x^3 + \sqrt{x^6 - 1} \right| + C \)
(c) \( \frac{1}{3} \log \left| x^3 - \sqrt{x^6 - 1} \right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{3} \log \left| x^3 + \sqrt{x^6 - 1} \right| + C \)
In simple words: Let \( t = x^3 \), so \( dt = 3x^2 dx \) and \( x^2 dx = \frac{dt}{3} \). The integral becomes \( \frac{1}{3} \int \frac{dt}{\sqrt{t^2 - 1}} = \frac{1}{3} \log \left| t + \sqrt{t^2 - 1} \right| + C \). Substitute back \( t = x^3 \) to get the final answer.

Exam Tip: Recognize that \( x^2 dx \) is part of the differential \( d(x^3) \), so substitution is the key strategy here.

Question 15. Mark (√) against the correct answer in each of the following: \( \int \frac{\sin x dx}{\sqrt{4\cos^2 x - 1}} = ? \)
(a) \( -\frac{1}{2} \log \left| 2\cos x + \sqrt{4\cos^2 x - 1} \right| + C \)
(b) \( -\frac{1}{3} \log \left| 2\cos x + \sqrt{4\cos^2 x - 1} \right| + C \)
(c) \( -\frac{1}{6} \log \left| 2\cos x + \sqrt{2\cos^2 x - 1} \right| + C \)
(d) None of the options
Answer: (a) \( -\frac{1}{2} \log \left| 2\cos x + \sqrt{4\cos^2 x - 1} \right| + C \)
In simple words: Let \( t = 2\cos x \), so \( dt = -2\sin x dx \) and \( \sin x dx = -\frac{dt}{2} \). The integral becomes \( -\frac{1}{2} \int \frac{dt}{\sqrt{t^2 - 1}} = -\frac{1}{2} \log \left| t + \sqrt{t^2 - 1} \right| + C \). Replace \( t \) with \( 2\cos x \).

Exam Tip: Notice that \( \sin x dx = -d(\cos x) \), so look for trigonometric substitutions when the integrand contains sine or cosine terms.

Question 16. Mark (√) against the correct answer in each of the following: \( \int \frac{\sec^2 x dx}{\sqrt{\tan^2 x - 4}} = ? \)
(a) \( \log \left| \tan x - \sqrt{\tan^2 x - 4} \right| + C \)
(b) \( \log \left| \tan x + \sqrt{\tan^2 x - 4} \right| + C \)
(c) \( \frac{1}{2} \log \left| \tan x + \sqrt{\tan^2 x - 4} \right| + C \)
(d) None of the options
Answer: (b) \( \log \left| \tan x + \sqrt{\tan^2 x - 4} \right| + C \)
In simple words: Use the substitution \( t = \tan x \), so \( dt = \sec^2 x dx \). The integral becomes \( \int \frac{dt}{\sqrt{t^2 - 4}} = \log \left| t + \sqrt{t^2 - 4} \right| + C \). Substitute back \( t = \tan x \).

Exam Tip: Recognize that \( \sec^2 x dx \) is the differential of \( \tan x \), making substitution straightforward.

Question 17. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{(1 - e^{2x})} = ? \)
(a) \( \log \left| e^x + \sqrt{e^{2x} - 1} \right| + C \)
(b) \( \log \left| e^{-x} + \sqrt{e^{-2x} - 1} \right| + C \)
(c) \( -\log \left| e^{-x} + \sqrt{e^{-2x} - 1} \right| + C \)
(d) None of the options
Answer: (c) \( -\log \left| e^{-x} + \sqrt{e^{-2x} - 1} \right| + C \)
In simple words: Using substitution \( t = 1 - e^{2x} \), simplify to get \( y = -\frac{1}{2}(-\log(1 - t) + \log t) + c \). After expanding using logarithm properties and substituting back \( t = 1 - e^{2x} \), the expression simplifies to the given answer form.

Exam Tip: For exponential integrals, careful algebraic manipulation of logarithmic expressions is needed to arrive at the simplest form.

Question 18. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x^2 - 3x + 2}} = ? \)
(a) \( \log \left| (x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2} \right| + C \)
(b) \( \log \left| x + \sqrt{x^2 - 3x + 2} \right| + C \)
(c) \( \log \left| x - \sqrt{x^2 - 3x + 2} \right| + C \)
(d) None of the options
Answer: (b) \( \log \left| x + \sqrt{x^2 - 3x + 2} \right| + C \)
In simple words: Complete the square: \( x^2 - 3x + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} \). Rewrite as \( (x - \frac{3}{2})^2 - (\frac{1}{2})^2 \). Apply the logarithmic integral formula with the appropriate substitution to get the final answer.

Exam Tip: Completing the square is the crucial first step for any quadratic expression under a square root sign.

Question 19. Mark (√) against the correct answer in each of the following: \( \int \frac{\cos x dx}{\sqrt{\sin^2 x - 2\sin x - 3}} = ? \)
(a) \( \log \left| \sin x + \sqrt{\sin^2 x - 2\sin x - 3} \right| + C \)
(b) \( \log \left| (\sin x - 1) + \sqrt{\sin^2 x - 2\sin x - 3} \right| + C \)
(c) \( \log \left| (\sin x - 1) - \sqrt{\sin^2 x - 2\sin x - 3} \right| + C \)
(d) None of the options
Answer: (b) \( \log \left| (\sin x - 1) + \sqrt{\sin^2 x - 2\sin x - 3} \right| + C \)
In simple words: Let \( t = \sin x \), so \( dt = \cos x dx \). The denominator becomes \( \sqrt{t^2 - 2t - 3} \). Complete the square: \( t^2 - 2t - 3 = (t - 1)^2 - 4 \). Apply the logarithmic integral formula to get \( \log \left| (t - 1) + \sqrt{(t - 1)^2 - 4} \right| + C \). Substitute back \( t = \sin x \).

Exam Tip: For integrals involving trigonometric functions in a denominator, always substitute to reduce it to a polynomial form first.

Question 20. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{2 - 4x + x^2}} = ? \)
(a) \( \log \left| (x - 2) + \sqrt{x^2 - 4x + 2} \right| + C \)
(b) \( \log \left| x + \sqrt{x^2 - 4x + 2} \right| + C \)
(c) \( \log \left| x - \sqrt{x^2 - 4x + 2} \right| + C \)
(d) None of the options
Answer: (b) \( \log \left| x - 2 + \sqrt{x^2 - 4x + 2} \right| + C \)
In simple words: Rearrange and complete the square: \( 2 - 4x + x^2 = (x - 2)^2 - 2 \). Rewrite as \( (x - 2)^2 - (\sqrt{2})^2 \). Apply the standard logarithmic integral formula to obtain the answer.

Exam Tip: Always arrange the quadratic in standard form \( ax^2 + bx + c \) before completing the square.

Question 21. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x^2 + 6x + 5}} = ? \)
(a) Generated content area (a)
(b) Generated content area (b)
(c) Generated content area (c)
(d) None of the options
Answer: Complete the square: \( x^2 + 6x + 5 = (x + 3)^2 - 4 \). Rewrite as \( (x + 3)^2 - 2^2 \). Using the standard formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C \), substitute to get \( \log \left| (x + 3) + \sqrt{x^2 + 6x + 5} \right| + C \).
In simple words: After completing the square and simplifying, the integral matches the standard logarithmic form and yields a logarithm of the transformed variable plus its associated square root.

Exam Tip: Verify your completed square form by expanding it to ensure no algebraic errors before applying the formula.

Question 22. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{(x - 3)^2 - 1}} = ? \)
(a) \( \log \left| (x - 3) + \sqrt{x^2 - 6x + 8} \right| + C \)
(b) \( \log \left| x + \sqrt{x^2 - 6x + 8} \right| + C \)
(c) \( \log \left| (x - 3) - \sqrt{x^2 - 6x + 8} \right| + C \)
(d) None of the options
Answer: (a) \( \log \left| (x - 3) + \sqrt{x^2 - 6x + 8} \right| + C \)
In simple words: The denominator is already in the form \( \sqrt{u^2 - 1} \) where \( u = x - 3 \). Expanding \( (x - 3)^2 - 1 = x^2 - 6x + 8 \), and applying the standard logarithmic formula gives the answer directly.

Exam Tip: When the squared term is already isolated, no additional completing the square is needed—apply the formula immediately.

Question 23. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{x^2 + 6x + 5}} = ? \)
(a) Generated option A
(b) Generated option B
(c) Generated option C
(d) None of the options
Answer: Complete the square: \( x^2 + 6x + 5 = (x + 3)^2 - 4 \). Express as \( (x + 3)^2 - 2^2 \). Use the standard integral formula for \( \sqrt{u^2 - a^2} \) to get \( \log \left| (x + 3) + \sqrt{x^2 + 6x + 5} \right| + C \).
In simple words: The completion of the square transforms the expression into a difference of squares, making the standard logarithmic formula applicable.

Exam Tip: Keep intermediate steps organized and verify factorizations to prevent sign errors during simplification.

Question 24. Mark (√) against the correct answer in each of the following: \( \int \frac{x^2 dx}{\sqrt{x^6 + a^6}} = ? \)
(a) \( \frac{1}{3} \log \left| x^6 + a^6 \right| + C \)
(b) \( \frac{1}{3} \tan^{-1} \left( \frac{x^3}{a^3} \right) + C \)
(c) \( \frac{1}{3} \log \left| x^3 + \sqrt{x^6 + a^6} \right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{3} \log \left| x^3 + \sqrt{x^6 + a^6} \right| + C \)
In simple words: Let \( t = x^3 \), so \( dt = 3x^2 dx \) and \( x^2 dx = \frac{dt}{3} \). The integral becomes \( \frac{1}{3} \int \frac{dt}{\sqrt{t^2 + a^6}} = \frac{1}{3} \log \left| t + \sqrt{t^2 + a^6} \right| + C \). Substitute back \( t = x^3 \).

Exam Tip: Always look for substitutions that transform the integrand into a standard form; here, \( x^2 dx \) being a multiple of \( d(x^3) \) is the key insight.

Question 25. Mark (√) against the correct answer in each of the following: \( \int \frac{\sec^2 x dx}{\sqrt{16 + \tan^2 x}} = ? \)
(a) \( \log \left| \tan x + \sqrt{\tan^2 x + 16} \right| + C \)
(b) \( \log \left| x + \sqrt{\tan^2 x + 16} \right| + C \)
(c) \( \log \left| \tan x - \sqrt{\tan^2 x + 16} \right| + C \)
(d) None of the options
Answer: (a) \( \log \left| \tan x + \sqrt{\tan^2 x + 16} \right| + C \)
In simple words: Let \( t = \tan x \), so \( dt = \sec^2 x dx \). The integral becomes \( \int \frac{dt}{\sqrt{t^2 + 16}} = \int \frac{dt}{\sqrt{t^2 + 4^2}} \), which matches the standard form \( \int \frac{du}{\sqrt{u^2 + a^2}} = \log \left| u + \sqrt{u^2 + a^2} \right| + C \). Substitute back \( t = \tan x \).

Exam Tip: When you see \( \sec^2 x \) in the numerator, always try \( t = \tan x \) as your substitution choice.

Question 26. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{3x^2 + 6x + 12}} = ? \)
(a) \( \log \left| (x + 1) + \sqrt{x^2 + 2x + 4} \right| + C \)
(b) \( \frac{1}{3} \log \left| (x + 1) + \sqrt{x^2 + 2x + 4} \right| + C \)
(c) \( \frac{1}{\sqrt{3}} \log \left| (x + 1) + \sqrt{x^2 + 2x + 4} \right| + C \)
(d) None of the options
Answer: (c) \( \frac{1}{\sqrt{3}} \log \left| (x + 1) + \sqrt{x^2 + 2x + 4} \right| + C \)
In simple words: Factor out 3 from the square root: \( \sqrt{3x^2 + 6x + 12} = \sqrt{3} \sqrt{x^2 + 2x + 4} \). The integral becomes \( \frac{1}{\sqrt{3}} \int \frac{dx}{\sqrt{x^2 + 2x + 4}} \). Complete the square in the denominator: \( x^2 + 2x + 4 = (x + 1)^2 + 3 \). Apply the logarithmic formula to finish.

Exam Tip: Always factor out constants from under a square root to simplify the integral before applying formulas.

Question 27. Mark (√) against the correct answer in each of the following: \( \int \frac{dx}{\sqrt{2x^2 + 4x + 6}} = ? \)
(a) \( \frac{1}{2} \log \left| (x + 1) + \sqrt{x^2 + 2x + 3} \right| + C \)
(b) \( \frac{1}{\sqrt{2}} \log \left| (x + 1) + \sqrt{x^2 + 2x + 3} \right| + C \)
(c) \( \frac{1}{\sqrt{2}} \log \left| x + \sqrt{x^2 + 2x + 3} \right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{\sqrt{2}} \log \left| (x + 1) + \sqrt{x^2 + 2x + 3} \right| + C \)
In simple words: Factor 2 from the square root: \( \sqrt{2x^2 + 4x + 6} = \sqrt{2} \sqrt{x^2 + 2x + 3} \). The integral becomes \( \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{x^2 + 2x + 3}} \). Complete the square: \( x^2 + 2x + 3 = (x + 1)^2 + 2 \). Apply the standard formula with the leading coefficient included.

Exam Tip: Factor out the coefficient of \( x^2 \) before completing the square to ensure accurate calculations.

Question 28. Mark (√) against the correct answer in each of the following: \( \int \frac{x^2 dx}{\sqrt{x^6 + 2x^3 + 3}} = ? \)
(a) \( \frac{1}{3} \log \left| (x^3 + 1) + \sqrt{x^6 + 2x^3 + 3} \right| + C \)
(b) \( \log \left| x^3 + \sqrt{x^6 + 2x^3 + 3} \right| + C \)
(c) \( \frac{1}{3} \log \left| (x^3 + 1) - \sqrt{x^6 + 2x^3 + 3} \right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{3} \log \left| (x^3 + 1) + \sqrt{x^6 + 2x^3 + 3} \right| + C \)
In simple words: Let \( t = x^3 \), so \( dt = 3x^2 dx \) and \( x^2 dx = \frac{dt}{3} \). The integral becomes \( \frac{1}{3} \int \frac{dt}{\sqrt{t^2 + 2t + 3}} \). Complete the square in the denominator: \( t^2 + 2t + 3 = (t + 1)^2 + 2 \). Apply the logarithmic formula and substitute back.

Exam Tip: When the integrand has a higher power of x in the numerator, look for a substitution that simplifies it to a lower degree polynomial.

Question 29. Mark (√) against the correct answer in each of the following: \( \int \sqrt{4 - x^2} dx = ? \)
(a) \( \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \frac{x}{2} + C \)
(b) Generated option B
(c) Generated option C
(d) None of the options
Answer: (a) \( \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \frac{x}{2} + C \)
In simple words: This is a standard form: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C \) with \( a = 2 \). Substituting gives the answer directly.

Exam Tip: Memorize the standard formula for integrals of the form \( \int \sqrt{a^2 - x^2} dx \) to solve these quickly.

Question 30. Mark (√) against the correct answer in each of the following: \( \int\sqrt{1 - 9x^2} \, dx = ? \)
(a) \( \frac{x}{2}\sqrt{1 - 9x^2} + \frac{1}{18}\sin^{-1} 3x + C \)
(b) \( \frac{3x}{2}\sqrt{1 - 9x^2} + \frac{1}{6}\sin^{-1} 3x + C \)
(c) \( \frac{x}{2}\sqrt{1 - 9x^2} + \frac{1}{6}\sin^{-1} 3x + C \)
(d) None of the options
Answer: (a) \( \frac{x}{2}\sqrt{1 - 9x^2} + \frac{1}{18}\sin^{-1} 3x + C \)
In simple words: Use the standard integral formula for \( \sqrt{a^2 - x^2} \) and substitute the appropriate values. The integral breaks into two parts - a square root term and an inverse sine term.

Exam Tip: Always identify the value of \( a \) from the expression under the square root and apply the standard formula carefully. Watch the coefficients in both terms of the final answer.

Question 31. Mark (√) against the correct answer in each of the following: \( \int\sqrt{9 - 4x^2} \, dx = ? \)
(a) \( \frac{x}{2}\sqrt{9 - 4x^2} + \frac{9}{4}\sin^{-1} \frac{2x}{3} + C \)
(b) \( x\sqrt{9 - 4x^2} + \frac{9}{2}\sin^{-1} \frac{2x}{3} + C \)
(c) \( \frac{x}{2}\sqrt{9 - 4x^2} - \frac{9}{4}\sin^{-1} \frac{2x}{3} + C \)
(d) None of the options
Answer: (a) \( \frac{x}{2}\sqrt{9 - 4x^2} + \frac{9}{4}\sin^{-1} \frac{2x}{3} + C \)
In simple words: The formula for \( \sqrt{a^2 - b^2x^2} \) yields two parts - one with the square root itself and one with an inverse sine. Match each coefficient correctly to your values.

Exam Tip: Rewrite \( 9 - 4x^2 \) as \( 3^2 - (2x)^2 \) to clearly identify \( a = 3 \) and the coefficient of \( x \). Double-check the fraction inside the inverse sine.

Question 32. Mark (√) against the correct answer in each of the following: \( \int\cos x\sqrt{9 - \sin^2 x} \, dx = ? \)
(a) \( \frac{1}{2}\sin x\sqrt{9 - \sin^2 x} + \frac{9}{2}\sin^{-1}\left( \frac{\sin x}{3} \right) + C \)
(b) \( \frac{\sin x}{2}\sqrt{9 - \sin^2 x} + \frac{9}{2}\sin^{-1}\left( \frac{\sin x}{3} \right) + C \)
(c) \( \frac{1}{2}\cos x\sqrt{9 - \sin^2 x} + \frac{9}{2}\sin^{-1}\left( \frac{\sin x}{3} \right) + C \)
(d) None of the options
Answer: (b) \( \frac{\sin x}{2}\sqrt{9 - \sin^2 x} + \frac{9}{2}\sin^{-1}\left( \frac{\sin x}{3} \right) + C \)
In simple words: Substitute \( \sin x = t \) so that \( \cos x \, dx = dt \). Then use the standard \( \sqrt{a^2 - t^2} \) formula with \( a = 3 \). At the end, replace \( t \) back with \( \sin x \).

Exam Tip: Substitution changes the integrand into a standard form. Always remember to replace the substitution variable back into the final answer before finishing.

Question 33. Mark (√) against the correct answer in each of the following: \( \int\sqrt{x^2 - 16} \, dx = ? \)
(a) \( x\sqrt{x^2 - 16} - 4\log\left|x + \sqrt{x^2 - 16}\right| + C \)
(b) \( \frac{x}{2}\sqrt{x^2 - 16} - 8\log\left|x + \sqrt{x^2 - 16}\right| + C \)
(c) \( \frac{x}{2}\sqrt{x^2 - 16} + 8\log\left|x + \sqrt{x^2 - 16}\right| + C \)
(d) None of the options
Answer: (b) \( \frac{x}{2}\sqrt{x^2 - 16} - 8\log\left|x + \sqrt{x^2 - 16}\right| + C \)
In simple words: For integrals of the form \( \sqrt{x^2 - a^2} \), apply the standard formula. Here \( a = 4 \), so \( a^2 = 16 \). The result has a square root term and a logarithm term with a minus sign.

Exam Tip: The sign in front of the logarithm term in the \( \sqrt{x^2 - a^2} \) formula is always negative. Be careful not to confuse it with the \( \sqrt{a^2 - x^2} \) case.

Question 34. Mark (√) against the correct answer in each of the following: \( \int\sqrt{x^2 - 4x + 2} \, dx = ? \)
(a) \( \frac{1}{2}(x - 2)\sqrt{x^2 - 4x + 2} + \log\left|(x - 2) + \sqrt{x^2 - 4x + 2}\right| + C \)
(b) \( (x - 2)\sqrt{x^2 - 4x + 2} + \frac{1}{2}\log\left|(x - 2) + \sqrt{x^2 - 4x + 2}\right| + C \)
(c) \( \frac{1}{2}(x - 2)\sqrt{x^2 - 4x + 2} - \log\left|(x - 2) + \sqrt{x^2 - 4x + 2}\right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{2}(x - 2)\sqrt{x^2 - 4x + 2} + \log\left|(x - 2) + \sqrt{x^2 - 4x + 2}\right| + C \)
In simple words: First, complete the square inside to get \( (x - 2)^2 - 2 \), which is of the form \( u^2 - a^2 \). Then apply the standard integral formula with \( u = x - 2 \) and \( a = \sqrt{2} \).

Exam Tip: Always complete the square for quadratic expressions under a square root. This transforms an unfamiliar form into a standard one that has a known formula.

Question 35. Mark (√) against the correct answer in each of the following: \( \int\sqrt{9x^2 + 16} \, dx = ? \)
(a) \( \frac{x}{2}\sqrt{9x^2 + 16} + \frac{8}{3}\log\left|3x + \sqrt{9x^2 + 16}\right| + C \)
(b) \( \frac{x}{2}\sqrt{9x^2 + 16} - \frac{8}{3}\log\left|3x + \sqrt{9x^2 + 16}\right| + C \)
(c) \( x\sqrt{9x^2 + 16} + 24\log\left|3x + \sqrt{9x^2 + 16}\right| + C \)
(d) None of the options
Answer: (a) \( \frac{x}{2}\sqrt{9x^2 + 16} + \frac{8}{3}\log\left|3x + \sqrt{9x^2 + 16}\right| + C \)
In simple words: Rewrite \( 9x^2 + 16 \) as \( (3x)^2 + 4^2 \) to match the standard form \( u^2 + a^2 \). Apply the integral formula with these substitutions, then simplify.

Exam Tip: When you see \( 9x^2 \), think of it as \( (3x)^2 \). This helps you identify the correct values to use in the standard formula and avoid calculation errors.

Question 36. Mark (√) against the correct answer in each of the following: \( \int e^x\sqrt{e^{2x} + 4} \, dx = ? \)
(a) \( \frac{1}{2}e^x\sqrt{e^{2x} + 4} - 2\log\left|e^x + \sqrt{e^{2x} + 4}\right| + C \)
(b) \( \frac{1}{2}e^x\sqrt{e^{2x} + 4} + 2\log\left|e^x + \sqrt{e^{2x} + 4}\right| + C \)
(c) \( e^x\sqrt{e^{2x} + 4} + \frac{1}{2}\log\left|e^x + \sqrt{e^{2x} + 4}\right| + C \)
(d) None of the options
Answer: (b) \( \frac{1}{2}e^x\sqrt{e^{2x} + 4} + 2\log\left|e^x + \sqrt{e^{2x} + 4}\right| + C \)
In simple words: Set \( e^x = t \), so \( e^x \, dx = dt \). The integral becomes \( \int\sqrt{t^2 + 4} \, dt \), which is a standard form with \( a = 2 \). Use that formula, then replace \( t \) with \( e^x \).

Exam Tip: Substitution converts this into a familiar problem. After applying the formula, always back-substitute to return to the original variable.

Question 37. Mark (√) against the correct answer in each of the following: \( \int\frac{\sqrt{16 + (\log x)^2}}{x} \, dx = ? \)
(a) \( \frac{1}{2}\log x \cdot\sqrt{16 + (\log x)^2} + 8\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + C \)
(b) \( \frac{1}{2}\log x \cdot\sqrt{16 + (\log x)^2} + 4\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + C \)
(c) \( \log x \cdot\sqrt{16 + (\log x)^2} + 16\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + C \)
(d) None of the options
Answer: (a) \( \frac{1}{2}\log x \cdot\sqrt{16 + (\log x)^2} + 8\log\left|\log x + \sqrt{16 + (\log x)^2}\right| + C \)
In simple words: Let \( \log x = t \), so \( \frac{1}{x} dx = dt \). The integral becomes \( \int\sqrt{16 + t^2} \, dt \), a standard form with \( a = 4 \). Apply the formula, then substitute back \( t = \log x \).

Exam Tip: Recognize that \( \frac{1}{x} dx \) is the derivative of \( \log x \). This is the key substitution that simplifies the integrand into a standard form.