RS Aggarwal Solutions for Class 12 Chapter 14 Some Special Integrals

Question 1. Evaluate: \( \int \frac{dx}{(1-9x)^2} \)
Answer: Let \( y = (1 - 9x) \). Differentiating with respect to \( x \), we find \( \frac{dy}{dx} = -9 \), which gives us \( dy = -9 \, dx \). Substituting into the integral:
\[ \int \frac{dx}{(1-9x)^2} = \int \frac{1}{y^2} \cdot \frac{dy}{-9} = -\frac{1}{9} \int y^{-2} \, dy = -\frac{1}{9} \cdot \frac{y^{-1}}{-1} + C = \frac{1}{9y} + C \]
Substituting back \( y = 1 - 9x \):
\[ \int \frac{dx}{(1-9x)^2} = \frac{1}{9(1-9x)} + C \]

Exam Tip: Always identify the inner function for substitution and compute its derivative carefully. Check your final answer by differentiating it back to match the original integrand.

Question 2. Evaluate: \( \int \frac{dx}{\sqrt{25-4x^2}} \)
Answer: Rewrite the denominator as \( \sqrt{25 - 4x^2} = \sqrt{4(\frac{25}{4} - x^2)} = 2\sqrt{(\frac{5}{2})^2 - x^2} \). Using the formula \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a}) + C \) with \( a = \frac{5}{2} \):
\[ \int \frac{dx}{\sqrt{25-4x^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{(\frac{5}{2})^2 - x^2} } = \frac{1}{2} \sin^{-1}\left(\frac{2x}{5}\right) + C \]

Exam Tip: Factor out constants from under the square root first, then apply the inverse sine formula. Verify your answer matches the form \( \sin^{-1}(...) \).

Question 3. Evaluate: \( \int \frac{dx}{\sqrt{x^2+16}} \)
Answer: Rewrite as \( \int \frac{dx}{\sqrt{4^2 + x^2}} \). Apply the formula \( \int \frac{dx}{\sqrt{a^2+x^2}} = \sinh^{-1}\left(\frac{x}{a}\right) + C \) or equivalently \( \ln\left|x + \sqrt{x^2+a^2}\right| + C \), where \( a = 4 \):
\[ \int \frac{dx}{\sqrt{x^2+16}} = \sinh^{-1}\left(\frac{x}{4}\right) + C \]
or
\[ = \ln\left|x + \sqrt{x^2+16}\right| + C \]

Exam Tip: Both forms are acceptable. The logarithmic form is often preferred when hyperbolic inverses have not been covered.

Question 4. Evaluate: \( \int \frac{dx}{\sqrt{4+9x^2}} \)
Answer: Factor the denominator: \( \sqrt{4 + 9x^2} = \sqrt{9(\frac{4}{9} + x^2)} = 3\sqrt{(\frac{2}{3})^2 + x^2} \). Using the formula \( \int \frac{dx}{\sqrt{a^2+x^2}} = \sinh^{-1}\left(\frac{x}{a}\right) + C \) with \( a = \frac{2}{3} \):
\[ \int \frac{dx}{\sqrt{4+9x^2}} = \frac{1}{3} \sinh^{-1}\left(\frac{3x}{2}\right) + C \]

Exam Tip: Recognize patterns where the denominator is a sum of squares. Factor out coefficients before applying the standard formula.

Question 5. Evaluate: \( \int \frac{dx}{\sqrt{50+2x^2}} \)
Answer: Rewrite as \( \int \frac{dx}{\sqrt{2(25 + x^2)}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{25 + x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{5^2+x^2}} \). Using the formula with \( a = 5 \):
\[ \int \frac{dx}{\sqrt{50+2x^2}} = \frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{x}{5}\right) + C = \frac{1}{10} \tan^{-1}\left(\frac{x}{5}\right) + C \]

Exam Tip: Always pull out constant factors from the radical before identifying the form of the integral.

Question 6. Evaluate: \( \int \frac{dx}{\sqrt{16x^2-25}} \)
Answer: Rewrite as \( \int \frac{dx}{\sqrt{(4x)^2 - 5^2}} \). Let \( y = x^3 \), so \( dy = 3x^2 \, dx \). This gives us:
\[ \int \frac{dx}{\sqrt{16x^2-25}} = \frac{1}{4} \int \frac{dx}{\sqrt{x^2 - (\frac{5}{4})^2}} = \frac{1}{4} \cosh^{-1}\left(\frac{4x}{5}\right) + C \]
or equivalently
\[ = \frac{1}{4} \ln\left|4x + \sqrt{16x^2-25}\right| + C \]

Exam Tip: For integrals with \( \sqrt{x^2 - a^2} \), use the hyperbolic cosine inverse or logarithmic form depending on the context.

Question 7. Evaluate: \( \int \frac{dx}{\sqrt{(x^2-1)}} \)
Answer: Recognize this as \( \int \frac{dx}{\sqrt{x^2 - 1^2}} \). Use the formula \( \int \frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}\left(\frac{x}{a}\right) + C \) or \( \ln\left|x + \sqrt{x^2-a^2}\right| + C \):
\[ \int \frac{dx}{\sqrt{x^2-1}} = \cosh^{-1}(x) + C = \ln\left|x + \sqrt{x^2-1}\right| + C \]

Exam Tip: This is a standard form. Memorize the difference between \( \sqrt{a^2-x^2} \) (inverse sine), \( \sqrt{a^2+x^2} \) (inverse sinh), and \( \sqrt{x^2-a^2} \) (inverse cosh).

Question 8. Evaluate: \( \int \frac{x^2}{(9+4x^2)} dx \)
Answer: Split the integrand: \( \frac{x^2}{9+4x^2} = \frac{1}{4} - \frac{9}{4(9+4x^2)} \). Thus:
\[ \int \frac{x^2}{(9+4x^2)} dx = \frac{1}{4} \int dx - \frac{9}{16} \int \frac{dx}{(x^2 + \frac{9}{4})} \]
Using \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C \) with \( a = \frac{3}{2} \):
\[ = \frac{x}{4} - \frac{9}{16} \cdot \frac{2}{3} \tan^{-1}\left(\frac{2x}{3}\right) + C = \frac{x}{4} - \frac{3}{8} \tan^{-1}\left(\frac{2x}{3}\right) + C \]

Exam Tip: Decompose rational functions where the degree of the numerator is less than the denominator by dividing out terms strategically.

Question 9. Evaluate: \( \int \frac{e^x}{(e^{2x}+1)} dx \)
Answer: Let \( y = e^x \), so \( dy = e^x \, dx \). The integral becomes:
\[ \int \frac{e^x}{e^{2x}+1} dx = \int \frac{dy}{y^2+1} = \tan^{-1}(y) + C = \tan^{-1}(e^x) + C \]

Exam Tip: Substitution is key here. Recognize that \( e^{2x} = (e^x)^2 \) and use the arctangent formula after substitution.

Question 10. Evaluate: \( \int \frac{\sin x}{(1+\cos^2 x)} dx \)
Answer: Let \( y = \cos x \), so \( dy = -\sin x \, dx \). Substituting:
\[ \int \frac{\sin x}{1+\cos^2 x} dx = \int \frac{-dy}{1+y^2} = -\tan^{-1}(y) + C = -\tan^{-1}(\cos x) + C \]

Exam Tip: Look for derivatives of common functions in the numerator. Here, \( \sin x \) is the derivative of \( -\cos x \), which guides the substitution.

Question 11. Evaluate: \( \int \frac{\cos x}{(1+\sin^2 x)} dx \)
Answer: Let \( y = \sin x \), so \( dy = \cos x \, dx \). The integral transforms to:
\[ \int \frac{\cos x}{1+\sin^2 x} dx = \int \frac{dy}{1+y^2} = \tan^{-1}(y) + C = \tan^{-1}(\sin x) + C \]

Exam Tip: This parallels the previous problem. Use the fact that \( \cos x \, dx \) is the differential of \( \sin x \).

Question 12. Evaluate: \( \int \frac{3x^5}{(1+x^{12})} dx \)
Answer: Let \( y = x^6 \), so \( dy = 6x^5 \, dx \), which means \( x^5 \, dx = \frac{dy}{6} \). Substituting:
\[ \int \frac{3x^5}{1+x^{12}} dx = \int \frac{3}{1+y^2} \cdot \frac{dy}{6} = \frac{1}{2} \int \frac{dy}{1+y^2} = \frac{1}{2} \tan^{-1}(y) + C = \frac{1}{2} \tan^{-1}(x^6) + C \]

Exam Tip: Recognize that \( x^{12} = (x^6)^2 \). The coefficient 3 in the numerator and the power \( x^5 \) indicate a substitution with \( y = x^6 \).

Question 13. Evaluate: \( \int \frac{2x^3}{(4+x^8)} dx \)
Answer: Let \( y = x^4 \), so \( dy = 4x^3 \, dx \), which gives us \( x^3 \, dx = \frac{dy}{4} \). Substituting:
\[ \int \frac{2x^3}{4+x^8} dx = \int \frac{2}{4+y^2} \cdot \frac{dy}{4} = \frac{1}{2} \int \frac{dy}{2^2 + y^2} = \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) + C = \frac{1}{4} \tan^{-1}\left(\frac{x^4}{2}\right) + C \]

Exam Tip: Notice that the exponent 8 is twice 4, and the coefficient 2 paired with \( x^3 \) suggests letting \( y = x^4 \).

Question 14. Evaluate: \( \int \frac{dx}{(e^x + e^{-x})} \)
Answer: Rewrite the denominator: \( e^x + e^{-x} = \frac{e^{2x}+1}{e^x} \). Thus:
\[ \int \frac{dx}{e^x + e^{-x}} = \int \frac{e^x}{e^{2x}+1} dx \]
Let \( y = e^x \), so \( dy = e^x \, dx \). Then:
\[ = \int \frac{dy}{y^2+1} = \tan^{-1}(y) + C = \tan^{-1}(e^x) + C \]

Exam Tip: Manipulate the denominator algebraically to recognize a standard form. This technique often simplifies apparently complex hyperbolic expressions.

Question 15. Evaluate: \( \int \frac{x}{(1-x^4)} dx \)
Answer: Let \( y = x^2 \), so \( dy = 2x \, dx \), which means \( x \, dx = \frac{dy}{2} \). Substituting:
\[ \int \frac{x}{1-x^4} dx = \int \frac{1}{1-y^2} \cdot \frac{dy}{2} = \frac{1}{2} \int \frac{dy}{1-y^2} \]
Using \( \int \frac{dy}{a^2-y^2} = \frac{1}{2a} \ln\left|\frac{a+y}{a-y}\right| + C \) with \( a = 1 \):
\[ = \frac{1}{2} \cdot \frac{1}{2} \ln\left|\frac{1+y}{1-y}\right| + C = \frac{1}{4} \ln\left|\frac{1+x^2}{1-x^2}\right| + C \]

Exam Tip: The form \( \frac{1}{a^2-b^2} \) signals the use of partial fractions or the logarithmic inverse hyperbolic form.

Question 16. Evaluate: \( \int \frac{x^2}{(a^6-x^6)} dx \)
Answer: Let \( y = x^3 \), so \( dy = 3x^2 \, dx \), which gives us \( x^2 \, dx = \frac{dy}{3} \). Substituting:
\[ \int \frac{x^2}{a^6-x^6} dx = \int \frac{1}{a^6-y^2} \cdot \frac{dy}{3} = \frac{1}{3} \int \frac{dy}{(a^3)^2 - y^2} \]
Using \( \int \frac{dy}{a^2-y^2} = \frac{1}{2a} \ln\left|\frac{a+y}{a-y}\right| + C \) with \( a = a^3 \):
\[ = \frac{1}{3} \cdot \frac{1}{2a^3} \ln\left|\frac{a^3+x^3}{a^3-x^3}\right| + C = \frac{1}{6a^3} \ln\left|\frac{a^3+x^3}{a^3-x^3}\right| + C \]

Exam Tip: Substitute so that \( x^6 = (x^3)^2 \) becomes apparent, then apply the standard difference-of-squares logarithmic formula.

Question 17. Evaluate: \( \int \frac{dx}{(x^2+4x+8)} \)
Answer: Complete the square in the denominator: \( x^2 + 4x + 8 = (x+2)^2 + 4 \). Let \( y = x + 2 \), so \( dy = dx \). Then:
\[ \int \frac{dx}{x^2+4x+8} = \int \frac{dy}{y^2+4} = \int \frac{dy}{y^2+2^2} = \frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) + C = \frac{1}{2} \tan^{-1}\left(\frac{x+2}{2}\right) + C \]

Exam Tip: Always complete the square first for quadratic denominators. This transforms them into the standard \( y^2 + a^2 \) form.

Question 18. Evaluate: \( \int \frac{dx}{(4x^2-4x+3)} \)
Answer: Factor out 4: \( 4x^2 - 4x + 3 = 4(x^2 - x) + 3 = 4((x - \frac{1}{2})^2 - \frac{1}{4}) + 3 = 4(x-\frac{1}{2})^2 + 2 \). Let \( y = x - \frac{1}{2} \), so \( dy = dx \). Then:
\[ \int \frac{dx}{4x^2-4x+3} = \int \frac{dy}{4y^2 + 2} = \frac{1}{4} \int \frac{dy}{y^2 + \frac{1}{2}} = \frac{1}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1}\left(\frac{y}{\frac{1}{\sqrt{2}}}\right) + C \]
\[ = \frac{\sqrt{2}}{4} \tan^{-1}(\sqrt{2} y) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\sqrt{2}(x - \frac{1}{2})\right) + C \]

Exam Tip: Completing the square with coefficients requires factoring out the leading coefficient first to avoid arithmetic errors.

Question 19. Evaluate: \( \int \frac{dx}{(2x^2+x+3)} \)
Answer: Factor out 2: \( 2x^2 + x + 3 = 2(x^2 + \frac{x}{2}) + 3 = 2((x+\frac{1}{4})^2 - \frac{1}{16}) + 3 = 2(x+\frac{1}{4})^2 + \frac{23}{8} \). Let \( y = x + \frac{1}{4} \), so \( dy = dx \). Then:
\[ \int \frac{dx}{2x^2+x+3} = \int \frac{dy}{2y^2 + \frac{23}{8}} = \frac{1}{2} \int \frac{dy}{y^2 + \frac{23}{16}} = \frac{1}{2} \cdot \frac{4}{\sqrt{23}} \tan^{-1}\left(\frac{4y}{\sqrt{23}}\right) + C \]
\[ = \frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right) + C \]

Exam Tip: After completing the square, factor out the coefficient of \( y^2 \) before applying the arctangent formula.

Question 20. Evaluate: \( \int \frac{dx}{(2x^2-x-1)} \)
Answer: Complete the square: \( 2x^2 - x - 1 = 2(x^2 - \frac{x}{2}) - 1 = 2((x-\frac{1}{4})^2 - \frac{1}{16}) - 1 = 2(x-\frac{1}{4})^2 - \frac{9}{8} \). Let \( y = x - \frac{1}{4} \), so \( dy = dx \). Then:
\[ \int \frac{dx}{2x^2-x-1} = \int \frac{dy}{2y^2 - \frac{9}{8}} = \frac{1}{2} \int \frac{dy}{y^2 - \frac{9}{16}} = \frac{1}{2} \cdot \frac{1}{2 \cdot \frac{3}{4}} \ln\left|\frac{y + \frac{3}{4}}{y - \frac{3}{4}}\right| + C \]
\[ = \frac{1}{3} \ln\left|\frac{4x+2}{4x-4}\right| + C = \frac{1}{3} \ln\left|\frac{2(x-1)}{2x+1}\right| + C \]

Exam Tip: When the completed square form gives a difference (not a sum), use the logarithmic formula for \( \int \frac{dy}{a^2-b^2} \).

Question 21. Evaluate: \( \int \frac{dx}{(3-2x-x^2)} \)
Answer: Rewrite as \( 3 - 2x - x^2 = -(x^2 + 2x - 3) = -((x+1)^2 - 4) = 4 - (x+1)^2 \). Let \( y = x + 1 \), so \( dy = dx \). Then:
\[ \int \frac{dx}{3-2x-x^2} = \int \frac{dy}{4-y^2} = \int \frac{dy}{2^2 - y^2} = \frac{1}{2 \cdot 2} \ln\left|\frac{2+y}{2-y}\right| + C = \frac{1}{4} \ln\left|\frac{x+3}{1-x}\right| + C \]

Exam Tip: When completing the square reveals a difference of squares, apply the logarithmic form immediately.

Question 22. Evaluate: \( \int \frac{x}{(x^2+3x+2)} dx \)
Answer: Factor the denominator: \( x^2 + 3x + 2 = (x+1)(x+2) \). Use partial fractions:
\[ \frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \]
Solving: \( x = A(x+2) + B(x+1) \). Setting \( x = -1 \) gives \( A = -1 \), and setting \( x = -2 \) gives \( B = 2 \). Thus:
\[ \int \frac{x}{x^2+3x+2} dx = \int \left(\frac{-1}{x+1} + \frac{2}{x+2}\right) dx = -\ln|x+1| + 2\ln|x+2| + C = \ln\left|\frac{(x+2)^2}{x+1}\right| + C \]

Exam Tip: Always factor quadratic denominators first. Partial fraction decomposition simplifies the integral into manageable logarithmic terms.

Question 23. Evaluate: \( \int \frac{(x-3)}{(x^2+2x-4)} dx \)
Answer: We apply partial fractions to break down the numerator. Let \( x - 3 = A(2x + 2) + B \). Equating coefficients of x gives \( 1 = 2A \), so \( A = \frac{1}{2} \). From the constant term, \( -3 = 2A + B \), we find \( B = -4 \). Substituting these values, the integral becomes \( \frac{1}{2} \log|x^2 + 2x - 4| - 4 \int \frac{1}{(x+1)^2 - 5} dx \). The second integral evaluates to \( \frac{1}{2\sqrt{5}} \log \left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| \). Combining both parts yields \( \frac{1}{2} \log|x^2 + 2x - 4| - \frac{2}{\sqrt{5}} \log \left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| + C \).
In simple words: Split the numerator using the partial fractions method. One part gives a simple logarithm, while the other requires completing the square to produce an inverse hyperbolic function or logarithm term.

Exam Tip: Always check if the numerator's degree is less than the denominator - if not, perform polynomial division first. Identifying the correct form of A and B is key to avoiding calculation errors.

Question 24. Evaluate: \( \int \frac{(2x-3)}{(x^2+3x-18)} dx \)
Answer: Using partial fractions, we express \( 2x - 3 = A(2x + 3) + B \). From the x coefficient, \( 2 = 2A \), giving \( A = 1 \). The constant term gives \( -3 = 3A + B \), so \( B = -6 \). The integral becomes \( \log|x^2 + 3x - 18| - 6 \int \frac{1}{(x+\frac{3}{2})^2 - (\frac{9}{2})^2} dx \). Completing the square and applying the standard formula for difference of squares yields \( \frac{1}{2 \cdot \frac{9}{2}} \log \left| \frac{x + \frac{3}{2} - \frac{9}{2}}{x + \frac{3}{2} + \frac{9}{2}} \right| = \frac{2}{9} \log \left| \frac{x - 3}{x + 6} \right| \). Thus, the complete result is \( \log|x^2 + 3x - 18| - \frac{2}{3} \log \left| \frac{x - 3}{x + 6} \right| + C \).
In simple words: Break the integrand into two parts using partial fractions. The first produces a standard logarithm of the denominator. The second requires completing the square to transform it into a recognizable form before integrating.

Exam Tip: Complete the square carefully and verify your transformation is correct before applying the integral formula. Double-check the final simplification of logarithms to match the given answer form.

Question 25. Evaluate: \( \int \frac{x^2}{(x^2+6x-3)} dx \)
Answer: Since the numerator degree equals the denominator degree, rewrite as \( \int \frac{x^2 + (6x - 3) - (6x - 3)}{(x^2 + 6x - 3)} dx = \int \frac{(x^2 + 6x - 3) - (6x - 3)}{(x^2 + 6x - 3)} dx = \int 1 dx - \int \frac{(6x - 3)}{(x^2 + 6x - 3)} dx \). The first integral yields x. For the second, use partial fractions with \( 6x - 3 = A(2x + 6) + B \). From coefficients, \( A = 3 \) and \( B = -21 \). This gives \( 3 \log|x^2 + 6x - 3| - 21 \int \frac{1}{(x+3)^2 - 12} dx \). Completing the square and applying the difference of squares formula produces \( \frac{21}{2\sqrt{12}} \log \left| \frac{x + 3 - 2\sqrt{3}}{x + 3 + 2\sqrt{3}} \right| = \frac{7\sqrt{3}}{4} \log \left| \frac{x + 3 - 2\sqrt{3}}{x + 3 + 2\sqrt{3}} \right| \). The final answer is \( x - 3 \log|x^2 + 6x - 3| + \frac{7\sqrt{3}}{4} \log \left| \frac{x + 3 - 2\sqrt{3}}{x + 3 + 2\sqrt{3}} \right| + C \).
In simple words: When the numerator and denominator have equal degree, divide to get a polynomial plus a fraction. Integrate the polynomial directly, then handle the remaining fraction via partial fractions and completing the square.

Exam Tip: Polynomial long division or careful algebraic manipulation is essential before applying partial fractions. Always verify that the numerator degree is strictly less than the denominator after rewriting.

Question 26. Evaluate: \( \int \frac{(2x-1)}{(2x^2+2x+1)} dx \)
Answer: Using partial fractions, express \( 2x - 1 = A(4x + 2) + B \). Equating coefficients of x yields \( 2 = 4A \), so \( A = \frac{1}{2} \). The constant term gives \( -1 = 2A + B \), hence \( B = -2 \). Substituting, the integral becomes \( \frac{1}{2} \log|2x^2 + 2x + 1| - 2 \int \frac{1}{2(x^2 + x + \frac{1}{2})} dx \). Completing the square inside, we get \( 2 \int \frac{1}{2((x + \frac{1}{2})^2 + \frac{1}{4})} dx = 2 \tan^{-1}(2x + 1) + C \). Therefore, the complete integral is \( \frac{1}{2} \log|2x^2 + 2x + 1| - 2 \tan^{-1}(2x + 1) + C \).
In simple words: Decompose the numerator using partial fractions. One part becomes a direct logarithm. The other leads to completing the square in the denominator, which transforms it into a form suitable for the inverse tangent integral.

Exam Tip: Recognize when completing the square yields a sum of squares (not a difference) - this signals an inverse tangent integral rather than a logarithm. Be precise with constants throughout the completion process.

Question 27. Evaluate: \( \int \frac{(1-3x)}{(3x^2+4x+2)} dx \)
Answer: Rewrite the integral by factoring out the negative: \( -\int \frac{(3x-1)}{(3x^2+4x+2)} dx \). Using partial fractions, set \( 3x - 1 = A(6x + 4) + B \). From the x coefficient, \( 3 = 6A \), giving \( A = \frac{1}{2} \). The constant yields \( -1 = 4A + B \), so \( B = -3 \). This splits the integral into \( -\frac{1}{2} \log|3x^2 + 4x + 2| + 3 \int \frac{1}{3(x^2 + \frac{4}{3}x + \frac{2}{3})} dx \). Completing the square gives \( x^2 + \frac{4}{3}x + \frac{2}{3} = (x + \frac{2}{3})^2 + \frac{2}{9} \). The remaining integral evaluates to \( \frac{\sqrt{2}}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}(x + \frac{2}{3})}{\sqrt{2}}\right) = \frac{\sqrt{2}}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}(3x+2)}{\sqrt{2}}\right) \). The final result is \( -\frac{1}{2} \log|3x^2 + 4x + 2| + \frac{\sqrt{2}}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}(3x+2)}{\sqrt{2}}\right) + C \).
In simple words: Factor out the negative sign first. Apply partial fractions to decompose the numerator. Complete the square to transform the denominator into a form that produces an inverse tangent integral combined with a logarithm.

Exam Tip: Pay close attention to signs when rewriting the integral. Completing the square accurately is crucial - a small error in the constant terms will propagate through the final answer.

Question 28. Evaluate: \( \int \frac{2x}{(2+x-x^2)} dx \)
Answer: Rewrite as \( -2 \int \frac{x}{(x^2-x-2)} dx \). Using partial fractions, set \( x = A(2x - 1) + B \). From the x coefficient, \( 1 = 2A \), so \( A = \frac{1}{2} \). The constant gives \( 0 = -A + B \), hence \( B = \frac{1}{2} \). The integral splits into \( -\log|x^2 - x - 2| - \frac{1}{2} \int \frac{1}{(x - \frac{1}{2})^2 - \frac{9}{4}} dx \). Completing the square yields \( -\frac{1}{2} \times \frac{1}{2 \times \frac{3}{2}} \log \left| \frac{x - 2}{x + 1} \right| = -\frac{1}{3} \log \left| \frac{x - 2}{x + 1} \right| \). Combining both parts, the result is \( -\log|x^2 - x - 2| + \frac{1}{3} \log \left| \frac{x - 2}{x + 1} \right| + C \), or equivalently \( -\log|2 + x - x^2| + \frac{1}{3} \log \left| \frac{1 + x}{2 - x} \right| + C \).
In simple words: Factor and rewrite to make partial fractions applicable. Split the integrand into parts, one giving a logarithm and the other requiring completing the square to produce another logarithm term after factoring.

Exam Tip: Always verify that completing the square produces a difference of squares (for logarithms) versus a sum of squares (for inverse trig). Sign changes in the denominator can flip between these cases.

Question 29. Evaluate: \( \int \frac{dx}{(1+\cos^2 x)} \)
Answer: Divide both numerator and denominator by \( \cos^2 x \) to obtain \( \int \frac{\sec^2 x dx}{1 + \sec^2 x} \). Let \( y = \tan x \), so \( dy = \sec^2 x dx \). Since \( \sec^2 x = y^2 + 1 \), the integral becomes \( \int \frac{dy}{1 + (y^2 + 1)} = \int \frac{dy}{y^2 + 2} \). This matches the standard form \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \) with \( a = \sqrt{2} \). Therefore, \( \int \frac{dy}{y^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{y}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C \).
In simple words: Divide by \( \cos^2 x \) to convert the integral into a form involving \( \sec^2 x \). Substitute \( y = \tan x \) to simplify, then use the standard inverse tangent formula.

Exam Tip: Dividing numerator and denominator by \( \cos^2 x \) is a key technique for trigonometric integrals. Always recognize when the result matches a standard form after substitution.

Question 30. Evaluate: \( \int \frac{dx}{(2+\sin^2 x)} \)
Answer: Divide numerator and denominator by \( \cos^2 x \) to get \( \int \frac{\sec^2 x dx}{2 \sec^2 x + \tan^2 x} \). Let \( y = \tan x \), giving \( dy = \sec^2 x dx \). Since \( \sec^2 x = y^2 + 1 \), the denominator becomes \( 2(y^2 + 1) + y^2 = 3y^2 + 2 \). The integral is \( \int \frac{dy}{3y^2 + 2} = \int \frac{dy}{(\sqrt{3}y)^2 + (\sqrt{2})^2} \). Using the standard formula with \( a = \sqrt{3}y \) and \( b = \sqrt{2} \), we get \( \frac{1}{\sqrt{3} \cdot \sqrt{2}} \tan^{-1}\left(\frac{\sqrt{3}y}{\sqrt{2}}\right) + C = \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{\sqrt{2}}\right) + C \).
In simple words: Transform the integral by dividing by \( \cos^2 x \). Substitute \( y = \tan x \) to reduce it to a recognizable form. Apply the inverse tangent formula with appropriate coefficients.

Exam Tip: After substitution, carefully identify the correct values for use in the standard formula. Coefficient errors are common - double-check the algebra inside the inverse tangent argument.

Question 31. Evaluate: \( \int \frac{dx}{(a^2\cos^2 x + b^2\sin^2 x)} \)
Answer: Divide numerator and denominator by \( \cos^2 x \) to obtain \( \int \frac{\sec^2 x dx}{a^2 + b^2 \tan^2 x} \). Let \( y = \tan x \), so \( dy = \sec^2 x dx \). The integral becomes \( \int \frac{dy}{a^2 + b^2 y^2} = \frac{1}{b^2} \int \frac{dy}{(\frac{a}{b})^2 + y^2} \). Using the standard formula, \( \frac{1}{b^2} \times \frac{1}{\frac{a}{b}} \tan^{-1}\left(\frac{y}{\frac{a}{b}}\right) + C = \frac{1}{ab} \tan^{-1}\left(\frac{b \tan x}{a}\right) + C \).
In simple words: Divide by \( \cos^2 x \) and substitute \( y = \tan x \). Factor out constants from the denominator to match the standard inverse tangent integral form.

Exam Tip: Keep track of the coefficients a and b throughout - they appear in both the denominator factorization and the final answer. A common error is mixing their positions in the inverse tangent argument.

Question 32. Evaluate: \( \int \frac{dx}{(\cos^2 x - 3\sin^2 x)} \)
Answer: Divide numerator and denominator by \( \cos^2 x \) to get \( \int \frac{\sec^2 x dx}{1 - 3 \tan^2 x} \). Let \( y = \tan x \), giving \( dy = \sec^2 x dx \). The integral becomes \( \int \frac{dy}{1 - 3y^2} = \frac{1}{3} \int \frac{dy}{(\frac{1}{\sqrt{3}})^2 - y^2} \). Using the standard formula for difference of squares, \( \frac{1}{3} \times \frac{1}{2 \times \frac{1}{\sqrt{3}}} \log \left| \frac{\frac{1}{\sqrt{3}} + y}{\frac{1}{\sqrt{3}} - y} \right| + C = \frac{1}{2\sqrt{3}} \log \left| \frac{1 + \sqrt{3} y}{1 - \sqrt{3} y} \right| + C \). Substituting \( y = \tan x \), the result is \( \frac{1}{2\sqrt{3}} \log \left| \frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x} \right| + C \).
In simple words: Divide by \( \cos^2 x \) and substitute \( y = \tan x \). Recognize that the denominator is a difference of squares, which yields a logarithmic integral rather than an inverse trig function.

Exam Tip: Distinguish between integrals of the form \( \int \frac{dy}{a^2 + y^2} \) (inverse trig) and \( \int \frac{dy}{a^2 - y^2} \) (logarithm). The sign in the denominator determines which formula applies.

Question 33. Evaluate: \( \int \frac{dx}{(\sin^2 x - 4\cos^2 x)} \)
Answer: Divide numerator and denominator by \( \cos^2 x \) to obtain \( \int \frac{\sec^2 x dx}{\tan^2 x - 4} \). Let \( y = \tan x \), so \( dy = \sec^2 x dx \). The integral becomes \( \int \frac{dy}{y^2 - 4} \). Using the formula for difference of squares, \( \int \frac{dy}{y^2 - 2^2} = \frac{1}{2 \times 2} \log \left| \frac{y - 2}{y + 2} \right| + C = \frac{1}{4} \log \left| \frac{\tan x - 2}{\tan x + 2} \right| + C \).
In simple words: Divide by \( \cos^2 x \) and use the substitution \( y = \tan x \). Apply the difference of squares logarithmic formula directly to the simplified denominator.

Exam Tip: Always check the sign and structure of the denominator after substitution to select the correct standard form - logarithm for differences, inverse trig for sums.

Question 34. Evaluate: \( \int \frac{dx}{(\sin x \cos x + 2\cos^2 x)} \)
Answer: Divide numerator and denominator by \( \cos^2 x \) to get \( \int \frac{\sec^2 x dx}{\tan x + 2} \). Let \( y = \tan x \), giving \( dy = \sec^2 x dx \). The integral becomes \( \int \frac{dy}{y + 2} = \log|y + 2| + C = \log|\tan x + 2| + C \).
In simple words: Factor the denominator and divide by \( \cos^2 x \). The resulting substitution \( y = \tan x \) simplifies the integral to a basic logarithmic form.

Exam Tip: When the denominator factors after dividing by \( \cos^2 x \), verify the factorization carefully - this often leads to a surprisingly simple integral.

Question 35. Evaluate: \( \int \frac{\sin 2x dx}{(\sin^4 x + \cos^4 x)} \)
Answer: Rewrite using \( \sin 2x = 2 \sin x \cos x \) to get \( \int \frac{2 \sin x \cos x dx}{(\sin^4 x + \cos^4 x)} \). Divide numerator and denominator by \( \cos^4 x \) to obtain \( \int \frac{2 \tan x \sec^2 x dx}{\tan^4 x + 1} \). Let \( y = \tan x \), so \( dy = \sec^2 x dx \). The integral becomes \( \int \frac{2y dy}{y^4 + 1} \). Now substitute \( z = y^2 \), giving \( dz = 2y dy \). This yields \( \int \frac{dz}{z^2 + 1} = \tan^{-1}(z) + C = \tan^{-1}(y^2) + C = \tan^{-1}(\tan^2 x) + C \).
In simple words: Use the double angle formula to rewrite the numerator. Divide by \( \cos^4 x \) and substitute \( y = \tan x \). A second substitution \( z = y^2 \) reduces the problem to the standard inverse tangent integral.

Exam Tip: Recognize when a composite substitution is needed - first change to trig functions via division, then make a u-substitution for the resulting polynomial integral.

Question 36. Evaluate: \( \int \frac{(2\sin 2\phi - \cos\phi)}{(6 - \cos^2\phi - 4\sin\phi)} d\phi \)
Answer: Rewrite the numerator: \( 2 \sin 2\phi - \cos\phi = 4 \sin\phi \cos\phi - \cos\phi = \cos\phi(4 \sin\phi - 1) \). Rewrite the denominator: \( 6 - \cos^2\phi - 4\sin\phi = 6 - (1 - \sin^2\phi) - 4\sin\phi = 5 + \sin^2\phi - 4\sin\phi \). Let \( y = \sin\phi \), so \( dy = \cos\phi d\phi \). The integral becomes \( \int \frac{(4y - 1)}{(y^2 - 4y + 5)} dy \). Using partial fractions, set \( 4y - 1 = A(2y - 4) + B \). From the y coefficient, \( 4 = 2A \), so \( A = 2 \). The constant gives \( -1 = -4A + B \), hence \( B = 7 \). This splits into \( 2 \log|y^2 - 4y + 5| + 7 \int \frac{1}{(y - 2)^2 + 1} dy = 2 \log|y^2 - 4y + 5| + 7 \tan^{-1}(y - 2) + C \). Substituting back, the result is \( 2 \log|\sin^2\phi - 4\sin\phi + 5| + 7 \tan^{-1}(\sin\phi - 2) + C \).
In simple words: Factor the numerator and simplify the denominator. Use the substitution \( y = \sin\phi \) to convert to a rational function. Apply partial fractions, completing the square where needed, then integrate using standard forms.

Exam Tip: When factoring numerators and denominators, look for common factors that cancel. Completing the square in the denominator is essential when the discriminant is negative (yielding inverse trig) or positive (yielding logarithms).

Question 37. Evaluate: \( \int \frac{dx}{(\sin x - 2\cos x)(2\sin x + \cos x)} \)
Answer: Using partial fraction decomposition, we express the integrand as:
\( \frac{1}{(\sin x - 2\cos x)(2\sin x + \cos x)} = \frac{A}{(\sin x - 2)} + \frac{B}{(2\sin x + 1)} \)
Multiply both sides by the denominator to get \( 1 = A(2\sin x + 1) + B(\sin x - 2) \).
When \( \sin x = 2 \): \( 1 = 5A \), so \( A = \frac{1}{5} \)
When \( \sin x = -\frac{1}{2} \): \( 1 = -\frac{5}{2}B \), so \( B = -\frac{2}{5} \)
Substituting back:
\( \int \frac{dx}{(\sin x - 2\cos x)(2\sin x + \cos x)} = \frac{1}{5}\ln|\sin x - 2| - \frac{1}{5}\ln|2\sin x + 1| + C \)
\( = \frac{1}{5}\log\left|\frac{\sin x - 2}{2\sin x + 1}\right| + C \)
In simple words: Split the fraction into simpler pieces using partial fractions, integrate each piece, then combine the results using logarithm properties.

Exam Tip: Always set up your partial fractions carefully - verify your constants A and B by substituting convenient values of the variable back into the equation.

Question 38. Evaluate: \( \int \frac{\left(1 - x^2\right)}{\left(1 + x^4\right)} dx \)
Answer: Divide the numerator and denominator by \( x^2 \):
\( \int \frac{\frac{1}{x^2} - 1}{\frac{1}{x^2} + x^2} dx \)
Let \( y = x + \frac{1}{x} \)
Then \( dy = \left(1 - \frac{1}{x^2}\right) dx \)
Notice that \( \frac{1}{x^2} - 1 = -\left(1 - \frac{1}{x^2}\right) = -dy \)
Also, \( \frac{1}{x^2} + x^2 = \left(x + \frac{1}{x}\right)^2 - 2 = y^2 - 2 \)
Therefore:
\( \int \frac{-dy}{y^2 - 2} = -\frac{1}{2\sqrt{2}}\ln\left|\frac{y - \sqrt{2}}{y + \sqrt{2}}\right| + C \)
Substituting \( y = x + \frac{1}{x} \):
\( = \frac{1}{2\sqrt{2}}\log\left|\frac{x + \frac{1}{x} + \sqrt{2}}{x + \frac{1}{x} - \sqrt{2}}\right| + C \)
\( = \frac{1}{2\sqrt{2}}\log\left|\frac{\sqrt{2x} + x^2 + 1}{\sqrt{2x} - x^2 + 1}\right| + C \)
In simple words: Use substitution to transform the integral into a standard logarithmic form, then work backwards to express the answer in terms of the original variable.

Exam Tip: When you see symmetric expressions like \( \frac{1}{x^2} + x^2 \), try dividing by the middle power and using a substitution that combines the variable with its reciprocal.

Question 39. Evaluate: \( \int \frac{\left(x^2 + 1\right)}{\left(x^4 + x^2 + 1\right)} dx \)
Answer: Divide both numerator and denominator by \( x^2 \):
\( \int \frac{1 + \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} dx \)
Let \( y = x - \frac{1}{x} \)
Then \( dy = \left(1 + \frac{1}{x^2}\right) dx \)
Also, \( x^2 + 1 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 3 = y^2 + 3 \)
Therefore:
\( \int \frac{dy}{y^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{y}{\sqrt{3}}\right) + C \)
Substituting \( y = x - \frac{1}{x} \):
\( = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2 - 1}{\sqrt{3}x}\right) + C \)
In simple words: Transform a complicated fraction by dividing top and bottom by \( x^2 \), use a clever substitution to get a standard inverse tangent form, then substitute back.

Exam Tip: When the denominator factors as a sum of squares after algebraic rearrangement, inverse trigonometric function integrals become your primary tool - watch for this pattern.

Question 40. Evaluate: \( \int \frac{dx}{\left(\sin^4 x + \cos^4 x\right)} \)
Answer: Divide the numerator and denominator by \( \cos^4 x \):
\( \int \frac{\sec^4 x}{\tan^4 x + 1} dx \)
\( = \int \frac{\sec^2 x(1 + \tan^2 x)}{(1 + \tan^4 x)} dx \)
Let \( y = \tan x \), so \( dy = \sec^2 x dx \)
Substituting:
\( \int \frac{1 + y^2}{1 + y^4} dy \)
Divide the numerator and denominator by \( y^2 \):
\( \int \frac{\frac{1}{y^2} + 1}{y^{-2} + y^2} dy \)
Let \( z = y - \frac{1}{y} \)
Then \( dz = \left(1 + \frac{1}{y^2}\right) dy \)
Also, \( y^{-2} + y^2 - 2 + 2 = \left(y - \frac{1}{y}\right)^2 + 2 = z^2 + 2 \)
Therefore:
\( \int \frac{dz}{z^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{z}{\sqrt{2}}\right) + C \)
Substituting back \( z = y - \frac{1}{y} = \tan x - \cot x \):
\( = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^2 x - 1}{\sqrt{2}\tan x}\right) + C \)
In simple words: Divide by the highest power of cosine, substitute for tangent, then divide again by a power and use another substitution to reach the inverse tangent form.

Exam Tip: When dealing with even powers of sine and cosine in the denominator, dividing by the appropriate power of cosine and converting to tangent simplifies the work significantly.

Question 1. Evaluate: \( \int \frac{dx}{\sqrt{16 - x^2}} \)
Answer: Use the standard formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{16 - x^2}} = \int \frac{dx}{\sqrt{4^2 - x^2}} = \sin^{-1}\left(\frac{x}{4}\right) + c \), where \( c \) is the integrating constant.
In simple words: Recognize that 16 equals \( 4^2 \), then directly apply the standard inverse sine formula.

Exam Tip: Always look for the difference of two squares under the square root sign - this signals the inverse sine formula; identify what \( a \) is immediately.

Question 2. Evaluate: \( \int \frac{dx}{\sqrt{1 - 9x^2}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{1 - 9x^2}} = \int \frac{dx}{\sqrt{9\left(\frac{1}{9} - x^2\right)}} = \frac{1}{3}\int \frac{dx}{\sqrt{\left(\frac{1}{3}\right)^2 - x^2}} = \frac{1}{3}\sin^{-1}(3x) + c \), where \( c \) is the integrating constant.
In simple words: Factor out 9 from the expression under the square root, then apply the standard formula with \( a = \frac{1}{3} \).

Exam Tip: When you have a coefficient in front of \( x^2 \), factor it out of the square root - this reveals the correct form for the standard formula.

Question 3. Evaluate: \( \int \frac{dx}{\sqrt{15 - 8x^2}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{15 - 8x^2}} = \int \frac{dx}{\sqrt{15\left(1 - \frac{8x^2}{15}\right)}} = \frac{1}{\sqrt{15}}\int \frac{dx}{\sqrt{1 - \left(\frac{\sqrt{8}}{\sqrt{15}}x\right)^2}} = \frac{1}{\sqrt{15}}\sin^{-1}\left(\frac{\sqrt{8}x}{\sqrt{15}}\right) + c \), where \( c \) is the integrating constant.
In simple words: Extract 15 from the square root as \( \sqrt{15} \), rewrite the remaining expression to match the standard form, then apply the inverse sine formula.

Exam Tip: Carefully match the coefficient of \( x^2 \) inside the square root to the correct inverse trigonometric form before applying the formula.

Question 4. Evaluate: \( \int \frac{dx}{\sqrt{x^2 - 4}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{x^2 - 4}} = \int \frac{dx}{\sqrt{x^2 - 2^2}} = \log\left|x + \sqrt{x^2 - 4}\right| + c \), where \( c \) is the integrating constant.
In simple words: Recognize \( x^2 - 4 \) as \( x^2 - 2^2 \) and apply the logarithmic formula for difference of squares under the square root.

Exam Tip: When you have a minus sign (difference) under the square root, use the logarithmic formula - this is distinct from the inverse sine case which uses addition.

Question 5. Evaluate: \( \int \frac{dx}{\sqrt{4x^2 - 1}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{4x^2 - 1}} = \int \frac{dx}{\sqrt{(2x)^2 - 1^2}} = \frac{1}{2}\log\left|2x + \sqrt{4x^2 - 1}\right| + c \), where \( c \) is the integrating constant.
In simple words: Factor the coefficient 4 to recognize \( (2x)^2 \), then apply the logarithmic formula and adjust for the substitution.

Exam Tip: When the coefficient of \( x^2 \) is not 1, complete a mental substitution to put it in standard form before applying the formula.

Question 6. Evaluate: \( \int \frac{dx}{\sqrt{9x^2 - 7}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{9x^2 - 7}} = \int \frac{dx}{\sqrt{(3x)^2 - \sqrt{7}^2}} = \log\left|3x + \sqrt{9x^2 - 7}\right| + c \), where \( c \) is the integrating constant.
In simple words: Write the expression as \( (3x)^2 - (\sqrt{7})^2 \), then use the logarithmic formula for difference of squares.

Exam Tip: Not every constant under the square root is a perfect square - recognize when you need to write it as a square root of another number.

Question 7. Evaluate: \( \int \frac{dx}{\sqrt{x^2 - 9}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{x^2 - 9}} = \int \frac{dx}{\sqrt{x^2 - 3^2}} = \log\left|x + \sqrt{x^2 - 9}\right| + c \), where \( c \) is the integrating constant.
In simple words: Identify 9 as \( 3^2 \) and directly apply the standard logarithmic formula.

Exam Tip: This is one of the most straightforward applications of the difference formula - ensure you don't miss the perfect square form.

Question 8. Evaluate: \( \int \frac{dx}{\sqrt{1 + 4x^2}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{1 + 4x^2}} = \int \frac{dx}{\sqrt{(2x)^2 + 1^2}} = \frac{1}{2}\log\left|2x + \sqrt{4x^2 + 1}\right| + c \), where \( c \) is the integrating constant.
In simple words: Rewrite the integrand as \( \sqrt{(2x)^2 + 1^2} \) and apply the formula for sum of squares, adjusting for the coefficient of \( x \).

Exam Tip: When you have a plus sign under the square root, use the logarithmic formula - be careful not to confuse this with the inverse sine formula which requires the form \( a^2 - x^2 \).

Question 9. Evaluate: \( \int \frac{dx}{\sqrt{9 + 4x^2}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{9 + 4x^2}} = \int \frac{dx}{\sqrt{(2x)^2 + 3^2}} = \frac{1}{2}\log\left|2x + \sqrt{4x^2 + 9}\right| + c \), where \( c \) is the integrating constant.
In simple words: Extract the coefficients as perfect squares - 9 becomes \( 3^2 \) and 4 becomes \( (2x)^2 \) - then apply the formula.

Exam Tip: Always extract coefficient factors from under the square root to match the standard form before integrating.

Question 10. Evaluate: \( \int \frac{x \, dx}{\sqrt{9 - x^4}} \)
Answer: Use the tip that \( d(x^2) = 2x \, dx \), so \( x \, dx = \frac{1}{2}d(x^2) \)
And the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{x \, dx}{\sqrt{9 - x^4}} = \frac{1}{2}\int \frac{d(x^2)}{\sqrt{3^2 - (x^2)^2}} = \frac{1}{2}\sin^{-1}\left(\frac{x^2}{3}\right) + c \), where \( c \) is the integrating constant.
In simple words: Treat \( x^2 \) as a single variable by using the differential \( d(x^2) = 2x \, dx \), then the integral matches the standard inverse sine pattern.

Exam Tip: When you see \( x \, dx \) in the numerator paired with a quartic expression in the denominator, use substitution to convert it into a simpler form.

Question 11. Evaluate: \( \int \frac{3x^2 \, dx}{\sqrt{9 - 16x^6}} \)
Answer: Use the tip that \( d(x^3) = 3x^2 \, dx \), so \( d(4x^3) = 4 \cdot 3x^2 \, dx \), which means \( 3x^2 \, dx = \frac{1}{4}d(2x^3) \)
And the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{3x^2 \, dx}{\sqrt{9 - 16x^6}} = \frac{1}{4}\int \frac{d(2x^3)}{\sqrt{3^2 - (4x^3)^2}} = \frac{1}{4}\sin^{-1}\left(\frac{4x^3}{3}\right) + c \), where \( c \) is the integrating constant.
In simple words: Recognize that \( 3x^2 \, dx \) is part of a derivative of a cubic power, use that substitution to simplify the denominator into standard form.

Exam Tip: High-degree expressions in the denominator often simplify through a carefully chosen substitution based on the numerator's differential structure.

Question 12. Evaluate: \( \int \frac{\sec^2 x \, dx}{\sqrt{16 + \tan^2 x}} \)
Answer: Use the tip that \( d(\tan x) = \sec^2 x \, dx \)
And the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{\sec^2 x \, dx}{\sqrt{16 + \tan^2 x}} = \int \frac{d(\tan x)}{\sqrt{4^2 + (\tan x)^2}} = \log\left|\tan x + \sqrt{16 + \tan^2 x}\right| + c \), where \( c \) is the integrating constant.
In simple words: Replace \( \sec^2 x \, dx \) with \( d(\tan x) \), then treat \( \tan x \) as the variable and apply the standard logarithmic formula.

Exam Tip: When trigonometric functions appear in denominators with their derivatives in numerators, immediately substitute to simplify the integral structure.

Question 13. Evaluate: \( \int \frac{\sin x \, dx}{\sqrt{4 + \cos^2 x}} \)
Answer: Use the tip that \( d(\cos x) = -\sin x \, dx \), so \( \sin x \, dx = -d(\cos x) \)
And the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{\sin x \, dx}{\sqrt{4 + \cos^2 x}} = \int \frac{-d(\cos x)}{\sqrt{(\cos x)^2 + 2^2}} = -\log\left|\cos x + \sqrt{4 + \cos^2 x}\right| + c \), where \( c \) is the integrating constant.
In simple words: Replace \( \sin x \, dx \) with \( -d(\cos x) \), treat \( \cos x \) as a new variable, and apply the standard formula.

Exam Tip: The negative sign arising from the derivative of cosine must be preserved throughout - it appears in the final answer as a negative coefficient.

Question 14. Evaluate: \( \int \frac{\cos x \, dx}{\sqrt{9\sin^2 x - 1}} \)
Answer: Use the tip that \( d(\sin x) = \cos x \, dx \), so \( d(3\sin x) = 3\cos x \, dx \), which means \( \cos x \, dx = \frac{1}{3}d(3\sin x) \)
And the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{\cos x \, dx}{\sqrt{9\sin^2 x - 1}} = \frac{1}{3}\int \frac{d(3\sin x)}{\sqrt{(3\sin x)^2 - 1^2}} = \frac{1}{3}\log\left|\cos x + \sqrt{4 + \cos^2 x}\right| + c \), where \( c \) is the integrating constant.
In simple words: Convert \( \cos x \, dx \) to a differential in terms of \( \sin x \), then use the logarithmic formula for the difference under the square root.

Exam Tip: Always check coefficients carefully when substituting - a coefficient in front of the trig function inside the square root requires a matching adjustment in the differential.

Question 15. Evaluate: \( \int \frac{e^x \, dx}{\sqrt{4 + e^{2x}}} \)
Answer: Use the tip that \( d(e^x) = e^x \, dx \)
And the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{e^x \, dx}{\sqrt{4 + e^{2x}}} = \int \frac{d(e^x)}{\sqrt{(e^x)^2 + 2^2}} = \log\left|e^x + \sqrt{4 + e^{2x}}\right| + c \), where \( c \) is the integrating constant.
In simple words: Substitute \( e^x \) as the variable by using \( d(e^x) = e^x \, dx \), then the integral takes the standard logarithmic form.

Exam Tip: Exponential and logarithmic integrals often become standard forms when you recognize that the numerator contains the derivative of the exponent.

Question 16. Evaluate: \( \int \frac{2e^x \, dx}{\sqrt{4 - e^{2x}}} \)
Answer: Use the tip that \( d(e^x) = e^x \, dx \)
And the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
\( \int \frac{2e^x \, dx}{\sqrt{4 - e^{2x}}} = 2\int \frac{d(e^x)}{\sqrt{2^2 - (e^x)^2}} = 2\sin^{-1}\left(\frac{e^x}{2}\right) + c \), where \( c \) is the integrating constant.
In simple words: Recognize \( 2e^x \, dx = 2d(e^x) \), then treat \( e^x \) as the integration variable and apply the inverse sine formula.

Exam Tip: With the minus sign under the square root, always switch to the inverse sine formula rather than the logarithmic one.

Question 17. Evaluate: \( \int \frac{dx}{\sqrt{1 - e^x}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{1 - e^x}} = \int \frac{dx}{\sqrt{e^x(e^{-x} - 1)}} = \int \frac{e^{-\frac{x}{2}} \, dx}{\sqrt{e^{-x} - 1}} \)
Let \( a = e^{-\frac{x}{2}} \), then \( -\frac{1}{2}e^{-\frac{x}{2}} \, dx = da \), so \( e^{-\frac{x}{2}} \, dx = -2da \)
\( = \int \frac{-2da}{\sqrt{a^2 - 1^2}} = -2\log\left|a + \sqrt{a^2 - 1}\right| + c = -2\log\left|e^{-\frac{x}{2}} + \sqrt{e^{-x} - 1}\right| + c \), where \( c \) is the integrating constant.
In simple words: Make a substitution to handle the exponential, then apply the standard logarithmic formula to the result.

Exam Tip: Exponentials in the denominator often require rearrangement and substitution before matching a standard form.

Question 18. Evaluate: \( \int \sqrt{\frac{a - x}{a + x}} \, dx \)
Answer: Use the substitution \( x = a\cos(2\theta) \)
\( dx = -2a\sin(2\theta) \, d\theta \) and \( \theta = \frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right) \)
\( \sin(2\theta) = \sqrt{1 - \frac{x^2}{a^2}} \)
\( \int \sqrt{\frac{a - x}{a + x}} \, dx = \int \sqrt{\frac{a - a\cos(2\theta)}{a + a\cos(2\theta)}} \times (-2a\sin(2\theta) \, d\theta) \)
\( = \int \sqrt{\frac{a(1 - \cos(2\theta))}{a(1 + \cos(2\theta))}} \times (-2a\sin(2\theta) \, d\theta) \)
Using \( \cos(2\theta) = 1 - 2\sin^2(\theta) \) and \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\( = -2a \int \frac{\sin(\theta)}{\cos(\theta)} \times 2\sin(\theta)\cos(\theta) \, d\theta = -2a \int 2\sin^2(\theta) \, d\theta = -2a \int (1 - \cos(2\theta)) \, d\theta \)
\( = -2a\left[\theta - \frac{\sin(2\theta)}{2}\right] + c = -2a\left[\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right) - \frac{\sin(2\theta)}{2}\right] + c \)
\( = -a\cos^{-1}\left(\frac{x}{a}\right) + a\sqrt{1 - \frac{x^2}{a^2}} + c = a\sin^{-1}\left(\frac{x}{a}\right) + \sqrt{a^2 - x^2} + c \), where \( c \) is the integrating constant.
In simple words: Use a trigonometric substitution to transform the square root into a simpler form, integrate using double angle formulas, then substitute back to the original variable.

Exam Tip: For square roots of rational expressions, trigonometric substitution paired with half-angle identities is often more efficient than algebraic manipulation alone.

Question 19. Evaluate: \( \int \frac{dx}{\sqrt{x^2 + 6x + 5}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
Complete the square in the expression under the radical:
\( x^2 + 6x + 5 = (x^2 + 2 \cdot x \cdot 3 + 3^2) + 5 - 9 = (x + 3)^2 - 4 = (x + 3)^2 - 2^2 \)
\( \int \frac{dx}{\sqrt{x^2 + 6x + 5}} = \int \frac{dx}{\sqrt{(x + 3)^2 - 2^2}} = \log\left|(x + 3) + \sqrt{x^2 + 6x + 5}\right| + c \), where \( c \) is the integrating constant.
In simple words: Complete the square to transform the quadratic into the form \( (ax + b)^2 - c^2 \), then apply the standard logarithmic formula.

Exam Tip: Always complete the square first when the denominator contains a quadratic expression - this converts it to a recognizable standard form.

Question 20. Evaluate: \( \int \frac{dx}{\sqrt{(2 - x)^2 + 1}} \)
Answer: Use the tip that \( d(2 - x) = -dx \), so \( dx = -d(2 - x) \)
And the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{(2 - x)^2 + 1}} = \int \frac{-d(2 - x)}{\sqrt{(2 - x)^2 + 1^2}} = -\log\left|(2 - x) + \sqrt{(2 - x)^2 + 1}\right| + c = -\log\left|(2 - x) + \sqrt{x^2 - 4x + 5}\right| + c \), where \( c \) is the integrating constant.
In simple words: Use substitution of the linear expression inside the square root, apply the standard formula, then substitute back.

Exam Tip: Linear transformations inside square roots (like \( 2 - x \)) should be converted immediately to differentials to simplify the integral.

Question 21. Evaluate: \( \int \frac{dx}{\sqrt{(x - 3)^2 + 1}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
\( \int \frac{dx}{\sqrt{(x - 3)^2 + 1}} = \log\left|(x - 3) + \sqrt{(x - 3)^2 + 1}\right| + c = \log\left|(x - 3) + \sqrt{x^2 - 6x + 10}\right| + c \), where \( c \) is the integrating constant.
In simple words: Recognize the sum of squares form \( (x - 3)^2 + 1^2 \) and apply the standard logarithmic formula directly.

Exam Tip: When the expression is already in the form of a perfect square plus a constant, no completing-the-square step is needed - apply the formula directly.

Question 22. Evaluate: \( \int \frac{dx}{\sqrt{x^2 - 6x + 10}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + c \) where \( c \) is the integrating constant.
Complete the square:
\( x^2 - 6x + 10 = (x^2 - 2 \cdot x \cdot 3 + 3^2) + 10 - 9 = (x - 3)^2 + 1 \)
\( \int \frac{dx}{\sqrt{x^2 - 6x + 10}} = \int \frac{dx}{\sqrt{(x - 3)^2 + 1}} = \log\left|(x - 3) + \sqrt{x^2 - 6x + 10}\right| + c \), where \( c \) is the integrating constant.
In simple words: Complete the square to get \( (x - 3)^2 + 1^2 \), then use the standard sum - of - squares formula.

Exam Tip: Completing the square is essential when the quadratic has a coefficient of 1 on \( x^2 \) but a non-zero linear term.

Question 23. Evaluate: \( \int \frac{dx}{\sqrt{2 + 2x - x^2}} \)
Answer: Use the formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + c \) where \( c \) is the integrating constant.
Complete the square:
\( 2 + 2x - x^2 = -(x^2 - 2x - 2) = -(x^2 - 2x + 1 - 1 - 2) = -(x^2 - 2x + 1 - 3) = 3 - (x - 1)^2 = (\sqrt{3})^2 - (x - 1)^2 \)
\( \int \frac{dx}{\sqrt{2 + 2x - x^2}} = \int \frac{dx}{\sqrt{(\sqrt{3})^2 - (x - 1)^2}} = \sin^{-1}\left(\frac{x - 1}{\sqrt{3}}\right) + c \), where \( c \) is the integrating constant.
In simple words: Factor out the negative sign from \( x^2 \), complete the square, express it as a difference of squares, then apply the inverse sine formula.

Exam Tip: When \( x^2 \) has a negative coefficient, you are dealing with a difference of squares under the square root - this signals the inverse sine formula, not the logarithmic one.

Question 24. Evaluate: \( \int \frac{dx}{\sqrt{8 - 4x - 2x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{8 - 4x - 2x^2}} \)

\( = \int \frac{dx}{\sqrt{10 - 2(x^2 + 2x + 1)}} \)

\( = \int \frac{dx}{\sqrt{(\sqrt{10})^2 - 2(x + 1)^2}} \)

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

\( = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{x+1}{\sqrt{5}}\right) + c \), c being the integration constant
In simple words: Rewrite the expression under the square root by completing the square, then factor out constants. This converts it into the standard form, allowing you to apply the inverse sine formula directly.

Exam Tip: Always complete the square first - this transforms the radicand into a difference of squares, which is essential for recognizing the correct formula to apply.

Question 25. Evaluate: \( \int \frac{dx}{\sqrt{16 - 6x - x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{16 - 6x - x^2}} \)

\( = \int \frac{dx}{\sqrt{25 - (x^2 + 6x + 9)}} \)

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

\( = \sin^{-1}\left(\frac{x+3}{5}\right) + c \), c being the integration constant
In simple words: Complete the square inside the square root to get a perfect square. Once you have the form \( a^2 - b^2 \), the inverse sine formula applies directly.

Exam Tip: Watch for the coefficient of \( x^2 \) - if it's not 1, factor it out before completing the square.

Question 26. Evaluate: \( \int \frac{dx}{\sqrt{7 - 6x - x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{7 - 6x - x^2}} \)

\( = \int \frac{dx}{\sqrt{16 - (x^2 + 6x + 9)}} \)

\( = \int \frac{dx}{\sqrt{(4)^2 - (x + 3)^2}} \)

\( = \sin^{-1}\left(\frac{x+3}{4}\right) + c \), c being the integration constant
In simple words: Rearrange the quadratic under the radical by completing the square to get \( a^2 - (x + b)^2 \). Then the inverse sine integration formula gives the answer immediately.

Exam Tip: Always ensure the radicand is in the form \( a^2 - [something]^2 \) before applying the inverse sine rule.

Question 27. Evaluate: \( \int \frac{dx}{\sqrt{x - x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{x - x^2}} \)

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

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

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

\( = \sin^{-1}(2x - 1) + c \), c being the integration constant
In simple words: Complete the square with \( x - x^2 \) to obtain \( \left(\frac{1}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2 \). Apply the inverse sine formula with \( a = \frac{1}{2} \).

Exam Tip: When the leading coefficient is negative, factor out the minus sign before completing the square.

Question 28. Evaluate: \( \int \frac{dx}{\sqrt{8 + 2x - x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{8 + 2x - x^2}} \)

\( = \int \frac{dx}{\sqrt{9 - (x^2 - 2x + 1)}} \)

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

\( = \sin^{-1}\left(\frac{x-1}{3}\right) + c \), c being the integration constant
In simple words: Finish the square to write \( 8 + 2x - x^2 \) as \( 9 - (x-1)^2 \). This puts it in the standard form for the inverse sine integral.

Exam Tip: Pay close attention to signs when collecting and rearranging terms under the square root.

Question 29. Evaluate: \( \int \frac{dx}{\sqrt{x^2 - 3x + 2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

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

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

\( = \int \frac{dx}{\sqrt{\left(x - \frac{3}{2}\right)^2 - \frac{1}{4}}} \)

\( = \log\left(x - \frac{3}{2}\right) + \sqrt{x^2 - 3x + 2} \mid + c \), c being the integration constant
In simple words: Complete the square to rewrite \( x^2 - 3x + 2 \) as \( \left(x - \frac{3}{2}\right)^2 - \frac{1}{4} \). This creates a difference of squares form, which matches the logarithmic integral formula.

Exam Tip: Remember: if the leading term is positive and you have a difference, use the logarithmic formula, not the inverse sine.

Question 30. Evaluate: \( \int \frac{dx}{\sqrt{2x^2 + 3x - 2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

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

\( = \int \frac{dx}{\sqrt{2\left(x^2 + 2 \times x \times \frac{3}{4} + \left(\frac{3}{4}\right)^2\right) - \frac{8}{8}}} \)

\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(x + \frac{3}{4}\right)^2 - \left(\frac{\sqrt{7}}{4}\right)^2}} \)

\( = \frac{1}{\sqrt{2}} \log\left[\left(x + \frac{3}{4}\right) + \sqrt{2x^2 + 3x - 2}\right] + c \), c being the integration constant
In simple words: Factor out the coefficient of \( x^2 \), then complete the square inside. The resulting form matches the logarithmic integral, which you can apply with the appropriate constant factor.

Exam Tip: Always extract coefficients of \( x^2 \) before completing the square - this simplifies the algebra significantly.

Question 31. Evaluate: \( \int \frac{dx}{\sqrt{2x^2 - 4x + 3}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

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

\( = \sin^{-1}\left(\frac{x-1}{3}\right) + c \), c being the integration constant
In simple words: Rework the expression under the square root by factoring and completing the square to obtain a difference-of-squares form. Then select and use the fitting inverse trigonometric integral formula.

Exam Tip: Identify whether your completed square gives \( a^2 - b^2 \) (use inverse sine) or \( b^2 - a^2 \) (use logarithm).

Question 32. Evaluate: \( \int \frac{dx}{\sqrt{1 + 2x - 3x^2}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

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

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

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

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

\( = \frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{x - \frac{1}{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\right) + c \)

\( = \frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{3x - 1}{\sqrt{2}}\right) + c \), c being the integration constant
In simple words: Factor out the coefficient of \( x^2 \) (with its negative sign), complete the square, and simplify the result. Then apply the inverse sine formula with the constant multiple in front.

Exam Tip: When the coefficient of \( x^2 \) is negative, factor it out carefully and adjust signs accordingly.

Question 33. Evaluate: \( \int \frac{dx}{\sqrt{2x^2 + 4x + 6}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{2x^2 + 4x + 6}} \)

\( = \int \frac{dx}{\sqrt{2(x^2 + 2x + 1) + 4}} \)

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

\( = \frac{1}{\sqrt{2}} \log\left[(x + 1) + \sqrt{2x^2 + 4x + 6}\right] + c \), c being the integration constant
In simple words: Factor out 2 from the expression, complete the square inside, and recognize the form \( u^2 + a^2 \). Apply the logarithmic integral formula with the constant factor in front.

Exam Tip: When you see a sum inside the square root (not a difference), use the logarithmic formula, not the inverse sine.

Question 34. Evaluate: \( \int \frac{dx}{\sqrt{x\sqrt{5} - x}} \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{dx}{\sqrt{5x - x^2}} \)

\( = \int \frac{dx}{\sqrt{\left(\frac{5}{2}\right)^2 - \left(x^2 - 2 \times x \times \frac{5}{2} + \left(\frac{5}{2}\right)^2\right)}} \)

\( = \int \frac{dx}{\sqrt{\left(\frac{5}{2}\right)^2 - \left(x - \frac{5}{2}\right)^2}} \)

\( = \sin^{-1}\left(\frac{x - \frac{5}{2}}{\frac{5}{2}}\right) + c \)

\( = \sin^{-1}\left(\frac{2x - 5}{5}\right) + c \), c being the integration constant
In simple words: Finish the square with \( 5x - x^2 \) to obtain \( \left(\frac{5}{2}\right)^2 - \left(x - \frac{5}{2}\right)^2 \). Then use the inverse sine formula directly.

Exam Tip: Always rearrange to get the form \( a^2 - (x - b)^2 \) before applying the inverse sine - this makes the formula clearer.

Question 35. Evaluate: \( \int \frac{x^2}{\sqrt{x^6 + 2x^3 + 3}} dx \)
Answer: Tip - \( d(x^3) = 3x^2 dx \), which means \( x^2 dx = \frac{1}{3} d(x^3) \)

The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \frac{x^2 dx}{\sqrt{x^6 + 2x^3 + 3}} \)

\( = \int \frac{\frac{1}{3} d(x^3)}{\sqrt{(x^3)^2 + 2x^3 + 3}} \)

\( = \frac{1}{3} \int \frac{d(x^3)}{\sqrt{(x^3 + 1)^2 + (\sqrt{2})^2}} \)

\( = \frac{1}{3} \log\left[(x^3 + 1) + \sqrt{x^6 + 2x^3 + 3}\right] + c \), c being the integration constant
In simple words: Identify that \( x^2 dx = \frac{1}{3} d(x^3) \) and substitute. Then complete the square in the new variable to apply the logarithmic integral formula.

Exam Tip: Look for patterns where the derivative of a composite function appears in the numerator - this signals a substitution opportunity.

Question 36. Evaluate: \( \int \frac{(2x + 3)}{\sqrt{x^2 + x + 1}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

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

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

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

Tip - Assuming \( x^2 + x + 1 = a^2 \), \( (2x + 1)dx = 2ada \)

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

\( = \int \frac{2ada}{a} \)

\( = \int 2da \)

\( = 2a + c_1 \)

\( = 2\sqrt{x^2 + x + 1} + c_1 \)

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

\( = 2\sqrt{x^2 + x + 1} + 2\log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x + 1}\right] + c \), c is the integration constant
In simple words: Split the numerator so that one part is the derivative of the expression under the square root. Handle each piece separately - one becomes \( 2\sqrt{x^2 + x + 1} \) and the other requires the logarithmic formula after completing the square.

Exam Tip: Break the numerator into two parts: a derivative of the denominator expression and a remainder. This divides a tough integral into two manageable pieces.

Question 37. Evaluate: \( \int \frac{(5x + 3)}{\sqrt{x^2 + 4x + 10}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \frac{(5x + 3)}{\sqrt{x^2 + 4x + 10}} dx \)

\( = \int \frac{\frac{5}{2} \times (2x + 4) - 7}{\sqrt{x^2 + 4x + 10}} dx \)

\( = \frac{5}{2} \int \frac{(2x + 4)}{\sqrt{x^2 + 4x + 10}} dx - \int \frac{7}{\sqrt{x^2 + 4x + 10}} dx \)

Tip - Assuming \( x^2 + 4x + 10 = a^2 \), \( (2x + 4)dx = 2ada \)

\( \frac{5}{2} \int \frac{(2x + 4)}{\sqrt{x^2 + 4x + 10}} dx \)

\( = \frac{5}{2} \int \frac{2ada}{a} \)

\( = \frac{5}{2} \int 2da \)

\( = 5a + c_1 \)

\( = 5\sqrt{x^2 + 4x + 10} + c_1 \)

\( \int \frac{7}{\sqrt{x^2 + 4x + 10}} dx \)

\( = 7 \int \frac{dx}{\sqrt{(x + 2)^2 + (\sqrt{6})^2}} \)

\( = 7\log\left[(x + 2) + \sqrt{x^2 + 4x + 10}\right] + c_2 \)

\( \int \frac{(5x + 3)}{\sqrt{x^2 + 4x + 10}} dx = 5\sqrt{x^2 + 4x + 10} - 7\log\left[(x + 2) + \sqrt{x^2 + 4x + 10}\right] + c \), c is the integration constant
In simple words: Rewrite the numerator as a multiple of the derivative of the radicand plus a constant. The first part integrates to a square root term; the second requires completing the square and using the logarithmic formula.

Exam Tip: When the numerator is linear, express it as \( A \cdot (derivative) + B \cdot (constant) \) to split the integral cleanly.

Question 38. Evaluate: \( \int \frac{(4x + 3)}{\sqrt{2x^2 + 2x - 3}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \frac{(4x + 3)}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \int \frac{(4x + 2) + 1}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \int \frac{(4x + 2)}{\sqrt{2x^2 + 2x - 3}} dx + \int \frac{1}{\sqrt{2x^2 + 2x - 3}} dx \)

Tip - Assuming \( 2x^2 + 2x - 3 = a^2 \), \( (4x + 2)dx = 2ada \)

\( \int \frac{(4x + 2)}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \int \frac{2ada}{a} \)

\( = \int 2da \)

\( = 2a + c_1 \)

\( = 2\sqrt{2x^2 + 2x - 3} + c_1 \)

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

\( = \int \frac{dx}{\sqrt{2\left(x + \frac{1}{2}\right)^2 - \left(\sqrt{\frac{7}{2}}\right)^2}} \)

\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\left(x + \frac{1}{2}\right)^2 - \left(\frac{\sqrt{7}}{2}\right)^2}} \)

\( = \frac{1}{\sqrt{2}} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x - \frac{3}{2}}\right] + c_2 \)

\( \int \frac{(4x + 3)}{\sqrt{2x^2 + 2x - 3}} dx + \int \frac{1}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = 2\sqrt{2x^2 + 2x - 3} + \frac{1}{\sqrt{2}} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x - \frac{3}{2}}\right] + c \), c is the integration constant
In simple words: Split the integral by writing the numerator as the derivative of the radicand plus a remainder. The first piece becomes \( 2\sqrt{2x^2 + 2x - 3} \); the second uses the logarithmic formula after completing the square.

Exam Tip: After completing the square in the denominator, check if you have \( a^2 - b^2 \) or \( b^2 - a^2 \) to pick the right integral formula.

Question 39. Evaluate: \( \int \frac{(3 - 2x)}{\sqrt{2 + x - x^2}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

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

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

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

Tip - Assuming \( 2 + x - x^2 = a^2 \), \( (1 - 2x)dx = 2ada \)

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

\( = \int \frac{2ada}{a} \)

\( = 2a + c_1 \)

\( = 2\sqrt{2 + x - x^2} + c_1 \)

\( \int \frac{2}{\sqrt{2 + x - x^2}} dx \)

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

\( = 2\sin^{-1}\left(\frac{x - \frac{1}{2}}{\frac{3}{2}}\right) + c_2 \)

\( = 2\sin^{-1}\left(\frac{2x - 1}{3}\right) + c_2 \)

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

\( = 2\sqrt{2 + x - x^2} + 2\sin^{-1}\left(\frac{2x - 1}{3}\right) + c \), c is the integration constant
In simple words: Break the numerator into the derivative of the radicand plus a constant term. The first part produces a square root; the second uses the inverse sine formula after completing the square.

Exam Tip: Always decompose the numerator to isolate the derivative part - this simplifies one integral immediately.

Question 40. Evaluate: \( \int \frac{(x + 2)}{\sqrt{2x^2 + 2x - 3}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

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

\( = \int \frac{\frac{1}{4} \times (4x + 2) + \frac{3}{2}}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \frac{1}{4} \int \frac{(4x + 2)}{\sqrt{2x^2 + 2x - 3}} dx + \frac{3}{2} \int \frac{1}{\sqrt{2x^2 + 2x - 3}} dx \)

Tip - Assuming \( 2x^2 + 2x - 3 = a^2 \), \( (4x + 2)dx = 2ada \)

\( \frac{1}{4} \int \frac{(4x + 2)}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \frac{1}{4} \int \frac{2ada}{a} \)

\( = \frac{1}{2} \int da \)

\( = \frac{a}{2} + c_1 \)

\( = \frac{\sqrt{2x^2 + 2x - 3}}{2} + c_1 \)

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

\( = \frac{3}{2} \int \frac{dx}{\sqrt{2\left(x + \frac{1}{2}\right)^2 - \left(\sqrt{\frac{7}{2}}\right)^2}} \)

\( = \frac{3}{2\sqrt{2}} \int \frac{dx}{\sqrt{\left(x + \frac{1}{2}\right)^2 - \left(\frac{\sqrt{7}}{2}\right)^2}} \)

\( = \frac{3}{2\sqrt{2}} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x - \frac{3}{2}}\right] + c_2 \)

\( \int \frac{(4x + 2)}{\sqrt{2x^2 + 2x - 3}} dx + \int \frac{1}{\sqrt{2x^2 + 2x - 3}} dx \)

\( = \frac{\sqrt{2x^2 + 2x - 3}}{2} + \frac{3}{2\sqrt{2}} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x - \frac{3}{2}}\right] + c \), c is the integration constant
In simple words: Rewrite the numerator as a coefficient times the derivative of the radicand, plus a remainder. Handle each piece separately to get a square root term and a logarithmic term.

Exam Tip: When the leading coefficient of the radicand is not 1, factor it out early to simplify all subsequent steps.

Question 41. Evaluate: \( \int \frac{(3x + 1)}{\sqrt{5 - 2x - x^2}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

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

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

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

Tip - Assuming \( 5 - 2x - x^2 = a^2 \), \( (-2 - 2x)dx = 2ada \), that is, \( (x + 1)dx = -ada \)

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

\( = -3 \int \frac{ada}{a} \)

\( = -3a + c_1 \)

\( = -3\sqrt{5 - 2x - x^2} + c_1 \)

\( \int \frac{2}{\sqrt{5 - 2x - x^2}} dx \)

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

\( = 2\sin^{-1}\left(\frac{x + 1}{\sqrt{6}}\right) + c_2 \)

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

\( = -3\sqrt{5 - 2x - x^2} - 2\sin^{-1}\left(\frac{x + 1}{\sqrt{6}}\right) + c \), c is the integration constant
In simple words: Factor the numerator so you can isolate a negative derivative of the radicand. One part then becomes a square root term with a minus sign; the other becomes an inverse sine term.

Exam Tip: Watch for negative derivatives - they produce negative signs in front of the square root term, so track the algebra carefully.

Question 42. Evaluate: \( \int \frac{(6x + 5)}{\sqrt{6 + x - 2x^2}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + c \) where c denotes the integration constant.

\( \int \frac{(6x + 5)}{\sqrt{6 + x - 2x^2}} dx \)

\( = \int \frac{\frac{3}{4}(4x - 1) + \frac{13}{2}}{\sqrt{6 + x - 2x^2}} dx \)

\( = \frac{3}{2} \int \frac{(4x - 1)}{\sqrt{6 + x - 2x^2}} dx + \frac{13}{2} \int \frac{1}{\sqrt{6 + x - 2x^2}} dx \)

Tip - Assuming \( 6 + x - 2x^2 = a^2 \), \( (1 - 4x)dx = 2ada \), that is, \( (4x - 1)dx = -2ada \)

\( \frac{3}{2} \int \frac{(4x - 1)}{\sqrt{6 + x - 2x^2}} dx \)

\( = -\frac{3}{2} \int \frac{2ada}{a} \)

\( = -3a + c_1 \)

\( = -3\sqrt{6 + x - 2x^2} + c_1 \)

\( \frac{13}{2} \int \frac{1}{\sqrt{6 + x - 2x^2}} dx \)

\( = \frac{13}{2} \int \frac{dx}{\sqrt{\left(\frac{7}{2\sqrt{2}}\right)^2 - 2\left(x - \frac{1}{4}\right)^2}} \)

\( = \frac{13}{2\sqrt{2}} \int \frac{dx}{\sqrt{\left(\frac{7}{4}\right)^2 - \left(x - \frac{1}{4}\right)^2}} \)

\( = \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{x - \frac{1}{4}}{\frac{7}{4}}\right) + c_2 \)

\( = \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{4x - 1}{7}\right) + c_2 \)

\( \int \frac{(4x - 1)}{\sqrt{6 + x - 2x^2}} dx + \int \frac{1}{\sqrt{6 + x - 2x^2}} dx \)

\( = -3\sqrt{6 + x - 2x^2} - 2\sin^{-1}\left(\frac{4x - 1}{7}\right) + c \), c is the integrating constant
In simple words: Express the numerator in terms of the derivative of the radicand plus a constant. Watch for negative derivatives that flip the sign. One piece produces a square root; the other uses the inverse sine formula.

Exam Tip: Always check the sign of the derivative relationship - a negative sign propagates through the whole term.

Question 43. Evaluate: \( \int \sqrt{\frac{1 + x}{x}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \sqrt{\frac{1 + x}{x}} dx \)

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

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

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

Tip - Taking \( x^2 + x = a^2 \), \( (2x + 1)dx = 2ada \)

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

\( = \frac{1}{2} \int \frac{2ada}{a} \)

\( = a + c_1 \)

\( = \sqrt{x^2 + x} + c_1 \)

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

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

\( = \frac{1}{2} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x}\right] + c_2 \)

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

\( = \sqrt{x^2 + x} + \frac{1}{2} \log\left[\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x}\right] + c \), c is the integration constant
In simple words: Simplify the radical fraction to get a rational form. Then express the numerator as a fraction of the derivative of the radicand plus a constant, and handle each piece separately.

Exam Tip: Simplifying radicals by multiplying numerator and denominator often reveals hidden structure that makes the substitution clearer.

Question 44. Evaluate: \( \int \frac{(x + 2)}{\sqrt{x^2 + 5x + 6}} dx \)
Answer: The formula being applied is \( \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \log(x + \sqrt{x^2 \pm a^2}) + c \) where c denotes the integration constant.

\( \int \frac{(x + 2)}{\sqrt{x^2 + 5x + 6}} dx \)

\( = \int \frac{\frac{1}{2}(2x + 5) - \frac{1}{2}}{\sqrt{x^2 + 5x + 6}} dx \)

\( = \frac{1}{2} \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx - \frac{1}{2} \int \frac{1}{\sqrt{x^2 + 5x + 6}} dx \)

Tip - Taking \( x^2 + 5x + 6 = a^2 \), \( (2x + 5)dx = 2ada \)

\( \frac{1}{2} \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx \)

\( = \frac{1}{2} \int \frac{2ada}{a} \)

\( = a + c_1 \)

\( = \sqrt{x^2 + 5x + 6} + c_1 \)

\( \frac{1}{2} \int \frac{1}{\sqrt{x^2 + 5x + 6}} dx \)

\( = \frac{1}{2} \int \frac{dx}{\sqrt{\left(x + \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} \)

\( = \frac{1}{2} \log\left[\left(x + \frac{5}{2}\right) + \sqrt{x^2 + 5x + 6}\right] + c_2 \)

\( \int \frac{2x + 5}{\sqrt{x^2 + 5x + 6}} dx - \int \frac{1}{\sqrt{x^2 + 5x + 6}} dx \)

\( = \sqrt{x^2 + 5x + 6} - \frac{1}{2} \log\left[\left(x + \frac{5}{2}\right) + \sqrt{x^2 + 5x + 6}\right] + c \), c is the integration constant
In simple words: Write the numerator as a fraction of the derivative of the radicand minus a remainder. One part gives a square root directly; the other requires completing the square and using the logarithmic formula.

Exam Tip: Always complete the square to identify the correct integration formula - verify you have the right form before applying the rule.

Exercise 14C

Question 1. Evaluate the following integrals: \( \int \sqrt{4 - x^2} dx \)
Answer: To Find: \( \int \sqrt{4 - x^2} dx \)

Now, \( \int \sqrt{4 - x^2} dx \) can be rewritten as \( \int \sqrt{2^2 - x^2} dx \)

Formula Used: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{2^2 - x^2} dx \) matches the form \( \int \sqrt{a^2 - x^2} dx \),

Hence, \( \int \sqrt{2^2 - x^2} dx = \frac{x}{2}\sqrt{2^2 - x^2} + \frac{2^2}{2} \sin^{-1}\frac{x}{2} + C \)

\( = \frac{x}{2}\sqrt{4 - x^2} + 2 \sin^{-1}\frac{x}{2} + C \)

Therefore, \( \int \sqrt{4 - x^2} dx = \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2} + C \)
In simple words: Recognize that \( 4 - x^2 = 2^2 - x^2 \). Apply the integration formula for \( \sqrt{a^2 - x^2} \) with \( a = 2 \) to get a combination of a square root term and an inverse sine term.

Exam Tip: Always express numbers under the square root as perfect squares (\( 4 = 2^2 \), \( 9 = 3^2 \), etc.) to match the standard formula quickly.

Question 2. Evaluate the following integrals: \( \int \sqrt{4 - 9x^2} dx \)
Answer: To Find: \( \int \sqrt{4 - 9x^2} dx \)

Now, \( \int \sqrt{4 - 9x^2} dx \) can be rewritten as \( \int \sqrt{2^2 - (3x)^2} dx \)

Formula Used: \( \int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{2^2 - (3x)^2} dx \) matches the form \( \int \sqrt{a^2 - x^2} dx \),
In simple words: Express \( 4 - 9x^2 \) as \( 2^2 - (3x)^2 \). Use a substitution or apply the formula with appropriate adjustment for the coefficient of \( x \).

Exam Tip: When the variable has a coefficient, factor it carefully before applying the standard integration formula.

Question 3. Evaluate the following integrals: \( \int \sqrt{x^2 - 2dx} \)
Answer: To find: \( \int \sqrt{x^2 - 2dx} \)

Rewrite \( \int \sqrt{x^2 - 2dx} \) as \( \int \sqrt{x^2 - (\sqrt{2})^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{x^2 - (\sqrt{2})^2 dx} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \),

Hence, \( \int \sqrt{x^2 - (\sqrt{2})^2 dx} = \frac{x}{2}\sqrt{x^2 - (\sqrt{2})^2} - \frac{(\sqrt{2})^2}{2} \log |x + \sqrt{x^2 - (\sqrt{2})^2}| + C \)

\( = \frac{x}{2}\sqrt{x^2 - 2} - \frac{2}{2} \log |x + \sqrt{x^2 - 2}| + C \)

\( = \frac{x}{2}\sqrt{x^2 - 2} - \log |x + \sqrt{x^2 - 2}| + C \)

Therefore, \( \int \sqrt{x^2 - 2dx} = \frac{x}{2}\sqrt{x^2 - 2} - \log |x + \sqrt{x^2 - 2}| + C \)
In simple words: Rewrite the expression under the square root as a difference of two squared terms. Then apply the standard integration formula for that form. The result combines a square root term with a logarithmic term.

Exam Tip: Always identify the constant term and rewrite it as a perfect square before applying the formula. Check that your final answer differentiates back to the original integrand.

Question 4. Evaluate the following integrals: \( \int \sqrt{2x^2 - 3dx} \)
Answer: To find: \( \int \sqrt{2x^2 - 3dx} \)

Rewrite \( \int \sqrt{2x^2 - 3 dx} \) as \( \int \sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2 dx} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \),

Hence, \( \int \sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2 dx} = \frac{\sqrt{2}x}{2}\sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2} - \frac{(\sqrt{3})^2}{2} \log |\sqrt{2}x + \sqrt{(\sqrt{2}x)^2 - (\sqrt{3})^2}| + C \)

\( = \frac{\sqrt{2}x}{2}\sqrt{2x^2 - 3} - \frac{3}{2} \log |\sqrt{2}x + \sqrt{2x^2 - 3}| + C \)

Therefore, \( \int \sqrt{2x^2 - 3dx} = \frac{\sqrt{2}x}{2}\sqrt{2x^2 - 3} - \frac{3}{2\sqrt{2}} \log |\sqrt{2}x + \sqrt{2x^2 - 3}| + C \)
In simple words: Factor out coefficients from the squared terms and rewrite the expression as a difference of squares. Apply the standard formula and simplify the result, keeping track of all constants throughout.

Exam Tip: When the coefficient of \( x^2 \) is not 1, factor it out completely before matching to the standard formula. Simplify logarithmic coefficients at the end.

Question 5. Evaluate the following integrals: \( \int \sqrt{x^2 + 5dx} \)
Answer: To find: \( \int \sqrt{x^2 + 5 dx} \)

Rewrite \( \int \sqrt{x^2 + 5 dx} \) as \( \int \sqrt{x^2 + (\sqrt{5})^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{x^2 + (\sqrt{5})^2 dx} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \),

Hence, \( \int \sqrt{x^2 + (\sqrt{5})^2 dx} = \frac{x}{2}\sqrt{x^2 + (\sqrt{5})^2} + \frac{(\sqrt{5})^2}{2} \log |x + \sqrt{x^2 + (\sqrt{5})^2}| + C \)

\( = \frac{x}{2}\sqrt{x^2 + 5} + \frac{5}{2} \log |x + \sqrt{x^2 + 5}| + C \)

Therefore, \( \int \sqrt{x^2 + 5 dx} = \frac{x}{2}\sqrt{x^2 + 5} + \frac{5}{2} \log |x + \sqrt{x^2 + 5}| + C \)
In simple words: Recognize the constant as a perfect square and apply the formula for a sum under the radical. The answer contains both a square root term and a logarithmic term added together.

Exam Tip: For sums (not differences), the logarithm term carries a plus sign rather than a minus sign. Always verify the form before selecting the correct formula.

Question 6. Evaluate the following integrals: \( \int \sqrt{4x^2 + 9dx} \)
Answer: To find: \( \int \sqrt{4x^2 + 9 dx} \)

Rewrite \( \int \sqrt{4x^2 + 9 dx} \) as \( \int \sqrt{(2x)^2 + 3^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{(2x)^2 + 3^2 dx} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \),

Hence, \( \int \sqrt{(2x)^2 + 3^2 dx} = \frac{2x}{2}\sqrt{(2x)^2 + 3^2} + \frac{3^2}{2} \log |2x + \sqrt{(2x)^2 + 3^2}| + C \)

\( = \frac{2x}{2}\sqrt{4x^2 + 9} + \frac{9}{2} \log |2x + \sqrt{4x^2 + 9}| + C \)

\( = x\sqrt{4x^2 + 9} + \frac{9}{2} \log |2x + \sqrt{4x^2 + 9}| + C \)

Therefore, \( \int \sqrt{4x^2 + 9dx} = x\sqrt{4x^2 + 9} + \frac{9}{2} \log |2x + \sqrt{4x^2 + 9}| + C \)
In simple words: Factor the coefficient from under the square root by extracting perfect squares. Match to the sum formula and simplify all terms, reducing fractions where possible.

Exam Tip: After applying the formula, simplify the coefficient fraction in front of the square root term. This often leads to a cleaner final form.

Question 7. Evaluate the following integrals: \( \int \sqrt{3x^2 + 4dx} \)
Answer: To find: \( \int \sqrt{3x^2 + 4 dx} \)

Rewrite \( \int \sqrt{3x^2 + 4 dx} \) as \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \),

Hence, \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} = \frac{\sqrt{3}x}{2}\sqrt{(\sqrt{3}x)^2 + 2^2} + \frac{2^2}{2} \log |\sqrt{3}x + \sqrt{(\sqrt{3}x)^2 + 2^2}| + C \)

\( = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + \frac{4}{2} \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)

\( = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + 2 \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)

Therefore, \( \int \sqrt{3x^2 + 4dx} = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + 2 \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)
In simple words: Extract the square root of the coefficient of \( x^2 \) to rewrite in standard form. Apply the formula and simplify, making sure the argument of the logarithm matches what appears under the radical.

Exam Tip: When extracting a square root from the coefficient, include it consistently in both the square root term and the logarithm argument.

Question 8. Evaluate the following integrals: \( \int \cos x\sqrt{9 - \sin^2 x dx} \)
Answer: To find: \( \int \cos x\sqrt{9 - \sin^2 x dx} \)

Let \( \sin x = t \)
\( \cos x dx = dt \)

Rewrite \( \int \cos x\sqrt{9 - \sin^2 x dx} \) as \( \int \sqrt{3^2 - t^2 dt} \)

Apply the formula: \( \int \sqrt{a^2 - x^2 dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{3^2 - t^2 dt} \) is in the form \( \int \sqrt{a^2 - x^2 dx} \) with \( t \) as the variable instead of \( x \),

\( \Rightarrow \int \sqrt{3^2 - t^2 dt} = \frac{1}{2}t\sqrt{3^2 - t^2} + \frac{3^2}{2} \sin^{-1}\frac{t}{3} + C \)

\( = \frac{t}{2}\sqrt{9 - t^2} + \frac{9}{2} \sin^{-1}\frac{t}{3} + C \)

Since \( \sin x = t \) and \( \cos x dx = dt \),

\( \Rightarrow \int \cos x\sqrt{9 - \sin^2 x dx} = \frac{\sin x}{2}\sqrt{9 - \sin^2 x} + \frac{9}{2} \sin^{-1}\left(\frac{\sin x}{3}\right) + C \)
In simple words: Substitute the sine function with a new variable. This transforms the integral into a standard form. After integrating, substitute back to return to the original variable.

Exam Tip: Substitution works best when the derivative of the substituted term appears in the integrand. Always substitute back at the end to match the original variable.

Question 9. Evaluate the following integrals: \( \int \sqrt{3x^2 + 4dx} \)
Answer: To find: \( \int \sqrt{3x^2 + 4 dx} \)

Rewrite \( \int \sqrt{3x^2 + 4 dx} \) as \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \),

Hence, \( \int \sqrt{(\sqrt{3}x)^2 + 2^2 dx} = \frac{\sqrt{3}x}{2}\sqrt{(\sqrt{3}x)^2 + 2^2} + \frac{2^2}{2} \log |\sqrt{3}x + \sqrt{(\sqrt{3}x)^2 + 2^2}| + C \)

\( = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + \frac{4}{2} \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)

\( = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + 2 \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)

Therefore, \( \int \sqrt{3x^2 + 4dx} = \frac{\sqrt{3}x}{2}\sqrt{3x^2 + 4} + 2 \log |\sqrt{3}x + \sqrt{3x^2 + 4}| + C \)
In simple words: Extract the square root of the leading coefficient and rewrite in the standard form. Apply the integration formula, then simplify all coefficients.

Exam Tip: Always rewrite the expression in standard form before applying the formula. Include the extracted coefficient consistently in both terms of the answer.

Question 10. Evaluate the following integrals: \( \int \sqrt{x^2 + 6x - 4dx} \)
Answer: To find: \( \int \sqrt{x^2 + 6x - 4 dx} \)

Rewrite \( \int \sqrt{x^2 + 6x - 4 dx} \) as \( \int \sqrt{x^2 + 6x + 3^2 - 3^2 - 4 dx} \)

i.e., \( \int \sqrt{(x + 3)^2 - 13 dx} \)

Let \( x + 3 = y \Rightarrow dx = dy \)

Rewrite \( \int \sqrt{(x + 3)^2 - 13 dx} \) as \( \int \sqrt{y^2 - (\sqrt{13})^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - (\sqrt{13})^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

\( \Rightarrow \int \sqrt{y^2 - (\sqrt{13})^2 dy} = \frac{y}{2}\sqrt{y^2 - (\sqrt{13})^2} - \frac{(\sqrt{13})^2}{2} \log |y + \sqrt{y^2 - (\sqrt{13})^2}| + C \)

\( = \frac{y}{2}\sqrt{y^2 - 13} - \frac{13}{2} \log |y + \sqrt{y^2 - 13}| + C \)

Since \( x + 3 = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{(x + 3)^2 - 13 dx} = \frac{(x+3)}{2}\sqrt{(x+3)^2 - 13} - \frac{13}{2} \log |(x + 3) + \sqrt{(x+3)^2 - 13}| + C \)

Therefore, \( \int \sqrt{x^2 + 6x - 4dx} = \frac{(x+3)}{2}\sqrt{x^2 + 6x - 4} - \frac{13}{2} \log |(x + 3) + \sqrt{x^2 + 6x - 4}| + C \)
In simple words: Complete the square in the expression under the radical. Substitute to simplify the quadratic into a difference of two squared terms. Then apply the appropriate formula.

Exam Tip: Completing the square is essential here. Make sure to add and subtract the same value to maintain equality, then factor the perfect square trinomial correctly.

Question 11. Evaluate the following integrals: \( \int \sqrt{2x - x^2 dx} \)
Answer: To find: \( \int \sqrt{2x - x^2 dx} \)

Rewrite \( \int \sqrt{2x - x^2 dx} \) as \( \int \sqrt{2x - x^2 - 1^2 + 1^2 dx} \)

i.e., \( \int \sqrt{1 - (x - 1)^2 dx} \)

Let \( x - 1 = y \Rightarrow dx = dy \)

Rewrite \( \int \sqrt{1 - (x - 1)^2 dx} \) as \( \int \sqrt{1^2 - y^2 dy} \)

Apply the formula: \( \int \sqrt{a^2 - x^2 dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{1^2 - y^2 dy} \) is in the form \( \int \sqrt{a^2 - x^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{1^2 - y^2 dy} = \frac{1}{2}y\sqrt{1^2 - y^2} + \frac{1^2}{2} \sin^{-1}\frac{y}{1} + C \)

\( = \frac{y}{2}\sqrt{1 - y^2} + \frac{1}{2} \sin^{-1}y + C \)

Since \( x - 1 = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{1 - (x - 1)^2 dx} = \frac{(x-1)}{2}\sqrt{1 - (x - 1)^2} + \frac{1}{2} \sin^{-1}(x - 1) + C \)

Therefore, \( \int \sqrt{2x - x^2 dx} = \frac{(x-1)}{2}\sqrt{2x - x^2} + \frac{1}{2} \sin^{-1}(x - 1) + C \)
In simple words: Rearrange the terms by completing the square to get a difference form. Substitute to convert into a standard format involving a sine inverse. Apply the formula and reverse the substitution.

Exam Tip: Completing the square transforms expressions like \( -x^2 + bx \) into perfect square forms. The inverse sine function appears when integrating expressions with a minus sign between squared terms.

Question 12. Evaluate the following integrals: \( \int \sqrt{1 - 4x - x^2 dx} \)
Answer: To find: \( \int \sqrt{1 - 4x - x^2 dx} \)

Rewrite \( \int \sqrt{1 - 4x - x^2 dx} \) as \( \int \sqrt{1 - 4x - x^2 - 2^2 + 2^2 dx} \)

i.e., \( \int \sqrt{5 - (x + 2)^2 dx} \)

Let \( x + 2 = y \Rightarrow dx = dy \)

Rewrite \( \int \sqrt{5 - (x + 2)^2 dx} \) as \( \int \sqrt{(\sqrt{5})^2 - y^2 dy} \)

Apply the formula: \( \int \sqrt{a^2 - x^2 dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{(\sqrt{5})^2 - y^2 dy} \) is in the form \( \int \sqrt{a^2 - x^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{(\sqrt{5})^2 - y^2 dy} = \frac{1}{2}y\sqrt{(\sqrt{5})^2 - y^2} + \frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{y}{\sqrt{5}} + C \)

\( = \frac{y}{2}\sqrt{5 - y^2} + \frac{5}{2} \sin^{-1}\frac{y}{\sqrt{5}} + C \)

Since \( x + 2 = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{5 - (x + 2)^2 dx} = \frac{(x+2)}{2}\sqrt{5 - (x + 2)^2} + \frac{5}{2} \sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \)

Therefore, \( \int \sqrt{1 - 4x - x^2 dx} = \frac{(x+2)}{2}\sqrt{1 - 4x - x^2} + \frac{5}{2} \sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C \)
In simple words: Complete the square to rewrite the expression in a difference form. Substitute the linear term to obtain a standard inverse sine form. Apply the formula and reverse the substitution at the end.

Exam Tip: When completing the square with a negative leading coefficient, factor out the negative first, then complete the square inside the parentheses. The constant term in the sine inverse argument must include the extracted coefficient.

Question 13. Evaluate the following integrals: \( \int \sqrt{2ax - x^2 dx} \)
Answer: To find: \( \int \sqrt{2ax - x^2 dx} \)

Rewrite \( \int \sqrt{2ax - x^2 dx} \) as \( \int \sqrt{2ax - x^2 - a^2 + a^2 dx} \)

i.e., \( \int \sqrt{a^2 - (x - a)^2 dx} \)

Let \( x - a = y \Rightarrow dx = dy \)

Rewrite \( \int \sqrt{a^2 - (x - a)^2 dx} \) as \( \int \sqrt{a^2 - y^2 dy} \)

Apply the formula: \( \int \sqrt{a^2 - x^2 dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{a^2 - y^2 dy} \) is in the form \( \int \sqrt{a^2 - x^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{a^2 - y^2 dy} = \frac{1}{2}y\sqrt{a^2 - y^2} + \frac{a^2}{2} \sin^{-1}\frac{y}{a} + C \)

\( = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2} \sin^{-1}\frac{y}{a} + C \)

Since \( x - a = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{a^2 - (x - a)^2 dx} = \frac{(x-a)}{2}\sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x-a}{a}\right) + C \)

Therefore, \( \int \sqrt{2ax - x^2 dx} = \frac{(x-a)}{2}\sqrt{2ax - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x-a}{a}\right) + C \)
In simple words: Complete the square by adding and subtracting \( a^2 \). Substitute the linear shift to achieve a standard difference form. Apply the inverse sine formula and back-substitute to return to the original variable.

Exam Tip: This is a parametric version where the constant \( a \) appears. Keep \( a \) as a symbol throughout and ensure it appears correctly in both the coefficient and the sine inverse argument.

Question 14. Evaluate the following integrals: \( \int \sqrt{2x^2 + 3x + 4dx} \)
Answer: To find: \( \int \sqrt{2x^2 + 3x + 4 dx} \)

Rewrite \( \int \sqrt{2x^2 + 3x + 4 dx} \) as \( \int \sqrt{2\left(x^2 + \frac{3}{2}x + 2\right)dx} \)

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

\( = \sqrt{2} \int \sqrt{x^2 + \frac{3}{2}x + \left(\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 + 2 dx} \)

\( = \sqrt{2} \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{23}{16} dx} \)

Let \( x + \frac{3}{4} = y \Rightarrow dx = dy \)

Rewrite as \( \sqrt{2} \int \sqrt{y^2 + \left(\frac{\sqrt{23}}{4}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{y^2 + \left(\frac{\sqrt{23}}{4}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \) with a change in variable,

\( \Rightarrow \int \sqrt{y^2 + \left(\frac{\sqrt{23}}{4}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 + \left(\frac{\sqrt{23}}{4}\right)^2} + \frac{\left(\frac{\sqrt{23}}{4}\right)^2}{2} \log |y + \sqrt{y^2 + \left(\frac{\sqrt{23}}{4}\right)^2}| + C \)

\( = \frac{y}{2}\sqrt{y^2 + \frac{23}{16}} + \frac{23}{32} \log |y + \sqrt{y^2 + \frac{23}{16}}| + C \)

Since \( x + \frac{3}{4} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{23}{16} dx} = \frac{1}{2}(4x + 3)\sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{23}{16}} + \frac{23}{32} \log \left|x + \frac{3}{4} + \sqrt{\left(x + \frac{3}{4}\right)^2 + \frac{23}{16}\right| + C \)

Now, \( \sqrt{2} \int \sqrt{x^2 + \frac{3}{2}x + 2 dx} = \frac{\sqrt{2}}{2}(4x + 3)\sqrt{2x^2 + 3x + 4} + \frac{23\sqrt{2}}{32} \log \left|x + \frac{3}{4} + \sqrt{2x^2 + 3x + 4}\right| + C \)

Therefore, \( \int \sqrt{2x^2 + 3x + 4dx} = \frac{1}{8}(4x + 3)\sqrt{2x^2 + 3x + 4} + \frac{23}{32} \log \left|\left(x + \frac{3}{4}\right) + \sqrt{2x^2 + 3x + 4}\right| + C \)
In simple words: Factor out the leading coefficient, complete the square inside, then substitute to reach standard form. Apply the formula and multiply back through by the extracted coefficient at the end.

Exam Tip: Keep careful track of coefficients when factoring and completing the square. The final constant in the logarithm may involve a square root and a fraction that simplify together.

Question 15. Evaluate the following integrals: \( \int \sqrt{x^2 + xdx} \)
Answer: To find: \( \int \sqrt{x^2 + x dx} \)

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

i.e., \( \int \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} dx} \)

Let \( x + \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

\( \Rightarrow \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 - \left(\frac{1}{2}\right)^2} - \frac{\left(\frac{1}{2}\right)^2}{2} \log \left|y + \sqrt{y^2 - \left(\frac{1}{2}\right)^2}\right| + C \)

\( = \frac{y}{2}\sqrt{y^2 - \frac{1}{4}} - \frac{1}{8} \log \left|y + \sqrt{y^2 - \frac{1}{4}}\right| + C \)

Since \( x + \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} dx} = \frac{1}{2}(2x + 1)\sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}} - \frac{1}{8} \log \left|\left(x + \frac{1}{2}\right) + \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right| + C \)

Therefore, \( \int \sqrt{x^2 + xdx} = \frac{1}{2}(2x + 1)\sqrt{x^2 + x} - \frac{1}{8} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| + C \)
In simple words: Complete the square by adding and subtracting the square of half the linear coefficient. Substitute to convert to difference form. Apply the standard formula and back-substitute.

Exam Tip: When completing the square, the coefficient in front of the square root term in the final answer equals 2 times the leading coefficient inside the integral. Simplify this product to get the cleanest form.

Question 16. Evaluate the following integrals: \( \int \sqrt{x^2 + xdx} \)
Answer: To find: \( \int \sqrt{x^2 + x dx} \)

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

i.e., \( \int \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} dx} \)

Let \( x + \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

\( \Rightarrow \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 - \left(\frac{1}{2}\right)^2} - \frac{\left(\frac{1}{2}\right)^2}{2} \log \left|y + \sqrt{y^2 - \left(\frac{1}{2}\right)^2}\right| + C \)

\( = \frac{y}{2}\sqrt{y^2 - \frac{1}{4}} - \frac{1}{8} \log \left|y + \sqrt{y^2 - \frac{1}{4}}\right| + C \)

Since \( x + \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} dx} = \frac{1}{2}(2x + 1)\sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}} - \frac{1}{8} \log \left|\left(x + \frac{1}{2}\right) + \sqrt{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right| + C \)

Therefore, \( \int \sqrt{x^2 + xdx} = \frac{1}{2}(2x + 1)\sqrt{x^2 + x} - \frac{1}{8} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| + C \)
In simple words: Complete the square in the radicand. Use substitution to simplify to a difference of squares. Apply the appropriate standard integral formula and reverse the substitution.

Exam Tip: The coefficient factor can be expanded and then simplified as \( 2x + 1 = 2(x + \frac{1}{2}) \). This matches the substitution and makes the answer cleaner.

Question 17. Evaluate the following integrals: \( \int (2x - 5)\sqrt{x^2 - 4x + 3dx} \)
Answer: To find: \( \int (2x - 5)\sqrt{x^2 - 4x + 3 dx} \)

Rewrite \( 2x - 5 \) as \( (2x - 4) - 1 \) and split the integral:

\( \int (2x - 5)\sqrt{x^2 - 4x + 3 dx} = \int [(2x - 4)\sqrt{x^2 - 4x + 3} - 1 \cdot \sqrt{x^2 - 4x + 3}]dx \)

\( = \int (2x - 4)\sqrt{x^2 - 4x + 3 dx} - \int \sqrt{x^2 - 4x + 3 dx} \)

For the first integral, let \( x^2 - 4x + 3 = u \Rightarrow dx = \frac{du}{2x-4} \)

Thus, \( \int (2x - 4)\sqrt{x^2 - 4x + 3 dx} = \int \sqrt{u} du = \frac{2}{3}u^{\frac{3}{2}} = \frac{2}{3}(x^2 - 4x + 3)^{\frac{3}{2}} \)

For the second integral, rewrite \( \int \sqrt{x^2 - 4x + 3 dx} = \int \sqrt{x^2 - 4x + 2^2 - 2^2 + 3 dx} = \int \sqrt{(x - 2)^2 - 1 dx} \)

Let \( x - 2 = y \Rightarrow dx = dy \)

Then \( \int \sqrt{(x - 2)^2 - 1 dx} = \int \sqrt{y^2 - 1^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - 1^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{y^2 - 1^2 dy} = \frac{y}{2}\sqrt{y^2 - 1} - \frac{1}{2} \log |y + \sqrt{y^2 - 1}| + C \)

Since \( x - 2 = y \) and \( dx = dy \),

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

Therefore, \( \int (2x - 4)\sqrt{x^2 - 4x + 3 dx} - \int \sqrt{x^2 - 4x + 3 dx} = \frac{2}{3}(x^2 - 4x + 3)^{\frac{3}{2}} - \frac{(x-2)}{2}\sqrt{x^2 - 4x + 3} + \frac{1}{2} \log |(x-2) + \sqrt{x^2 - 4x + 3}| + C \)
In simple words: Split the linear coefficient into a piece that matches the derivative of the radicand and another constant. The first piece integrates via substitution using the power rule. The second piece requires completing the square and applying the difference formula.

Exam Tip: Always decompose a linear coefficient so one part is the derivative of the expression under the radical. This avoids unnecessary complexity and leads to a cleaner solution.

Question 18. Evaluate the following integrals: \( \int (x + 2)\sqrt{x^2 + x + 1dx} \)
Answer: To find: \( \int (x + 2)\sqrt{x^2 + x + 1 dx} \)

Rewrite \( x + 2 \) as \( \frac{1}{2}(2x + 1) + \frac{3}{2} \) and split:

\( \int (x + 2)\sqrt{x^2 + x + 1 dx} = \int \left[\frac{1}{2}(2x + 1) + \frac{3}{2}\right]\sqrt{x^2 + x + 1 dx} \)

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

For the first integral, let \( x^2 + x + 1 = u \Rightarrow dx = \frac{du}{2x+1} \)

Thus, \( \frac{1}{2}\int (2x + 1)\sqrt{x^2 + x + 1 dx} = \frac{1}{2}\int \sqrt{u} du = \frac{1}{2} \cdot \frac{2}{3}u^{\frac{3}{2}} = \frac{1}{3}(x^2 + x + 1)^{\frac{3}{2}} \)

For the second integral, complete the square:

\( \int \sqrt{x^2 + x + 1 dx} = \int \sqrt{x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 dx} = \int \sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4} dx} \)

Let \( x + \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{y^2 + \left(\frac{\sqrt{3}}{2}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{y^2 + \left(\frac{\sqrt{3}}{2}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 + a^2 dx} \) with a change in variable,

\( \Rightarrow \int \sqrt{y^2 + \left(\frac{\sqrt{3}}{2}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 + \left(\frac{\sqrt{3}}{2}\right)^2} + \frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2} \log |y + \sqrt{y^2 + \left(\frac{\sqrt{3}}{2}\right)^2}| + C \)

\( = \frac{y}{2}\sqrt{y^2 + \frac{3}{4}} + \frac{3}{8} \log |y + \sqrt{y^2 + \frac{3}{4}}| + C \)

Since \( x + \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \frac{3}{2}\int \sqrt{x^2 + x + 1 dx} = \frac{3}{2}\left[\frac{1}{2}(2x + 1)\sqrt{x^2 + x + 1} + \frac{3}{8} \log \left|\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x + 1}\right|\right] + C \)

\( = \frac{3}{4}(2x + 1)\sqrt{x^2 + x + 1} + \frac{9}{16} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right| + C \)

Therefore, \( \int (x + 2)\sqrt{x^2 + x + 1 dx} = \frac{1}{3}(x^2 + x + 1)^{\frac{3}{2}} + \frac{3}{4}(2x + 1)\sqrt{x^2 + x + 1} + \frac{9}{16} \log \left|\left(x + \frac{1}{2}\right) + \sqrt{x^2 + x + 1}\right| + C \)
In simple words: Decompose the linear term so that one part is the derivative of the quadratic. The first part integrates directly using substitution. The second part requires completing the square and applying the standard sum formula.

Exam Tip: When splitting a linear coefficient, ensure the parts add back to the original. Keep track of all constant multiples through both integrals and combine them at the end.

Question 19. Evaluate the following integrals: \( \int (x - 5)\sqrt{x^2 + x dx} \)
Answer: To find: \( \int (x - 5)\sqrt{x^2 + x dx} \)

Rewrite \( x - 5 \) as \( \frac{1}{2}(2x + 1) - \frac{11}{2} \) and split:

\( \int (x - 5)\sqrt{x^2 + x dx} = \int \left[\frac{1}{2}(2x + 1) - \frac{11}{2}\right]\sqrt{x^2 + x dx} \)

\( = \frac{1}{2}\int (2x + 1)\sqrt{x^2 + x dx} - \frac{11}{2}\int \sqrt{x^2 + x dx} \)

For the first integral, let \( x^2 + x = u \Rightarrow dx = \frac{du}{2x+1} \)

Thus, \( \frac{1}{2}\int (2x + 1)\sqrt{x^2 + x dx} = \frac{1}{2}\int \sqrt{u} du = \frac{1}{2} \cdot \frac{2}{3}u^{\frac{3}{2}} = \frac{1}{3}(x^2 + x)^{\frac{3}{2}} \)

For the second integral, complete the square:

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

Let \( x + \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{y^2 - \left(\frac{1}{2}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 - \left(\frac{1}{2}\right)^2} - \frac{\left(\frac{1}{2}\right)^2}{2} \log |y + \sqrt{y^2 - \left(\frac{1}{2}\right)^2}| + C \)

\( = \frac{y}{2}\sqrt{y^2 - \frac{1}{4}} - \frac{1}{8} \log |y + \sqrt{y^2 - \frac{1}{4}}| + C \)

Since \( x + \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{x^2 + x dx} = \frac{1}{2}(2x + 1)\sqrt{x^2 + x} - \frac{1}{8} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| + C \)

Now,
\( - \frac{11}{2}\int \sqrt{x^2 + x dx} = - \frac{11}{2}\left[\frac{1}{2}(2x + 1)\sqrt{x^2 + x} - \frac{1}{8} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right|\right] + C \)

\( = - \frac{11}{4}(2x + 1)\sqrt{x^2 + x} + \frac{11}{16} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| + C \)

Therefore, \( \int (x - 5)\sqrt{x^2 + x dx} = \frac{1}{3}(x^2 + x)^{\frac{3}{2}} - \frac{11}{4}(2x + 1)\sqrt{x^2 + x} + \frac{11}{16} \log \left|x + \frac{1}{2} + \sqrt{x^2 + x}\right| + C \)
In simple words: Split the linear term into a part proportional to the derivative of the radicand and a constant part. Apply substitution to the first part. Complete the square and use the difference formula for the second part.

Exam Tip: When decomposing \( x - 5 \) as \( \frac{1}{2}(2x + 1) + k \), solve for \( k \) algebraically: \( x - 5 = \frac{1}{2}(2x + 1) + k \Rightarrow k = x - 5 - x - \frac{1}{2} = -\frac{11}{2} \).

Question 20. Evaluate the following integrals: \( \int (4x + 1)\sqrt{x^2 - x - 2dx} \)
Answer: To find: \( \int (4x + 1)\sqrt{x^2 - x - 2 dx} \)

Rewrite \( 4x + 1 \) as \( 2(2x - 1) + 3 \) and split:

\( \int (4x + 1)\sqrt{x^2 - x - 2 dx} = \int [2(2x - 1) + 3]\sqrt{x^2 - x - 2 dx} \)

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

For the first integral, let \( x^2 - x - 2 = u \Rightarrow dx = \frac{du}{2x-1} \)

Thus, \( 2\int (2x - 1)\sqrt{x^2 - x - 2 dx} = 2\int \sqrt{u} du = 2 \cdot \frac{2}{3}u^{\frac{3}{2}} = \frac{4}{3}(x^2 - x - 2)^{\frac{3}{2}} \)

For the second integral, complete the square:

\( \int \sqrt{x^2 - x - 2 dx} = \int \sqrt{x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 2 dx} = \int \sqrt{\left(x - \frac{1}{2}\right)^2 - \frac{9}{4} dx} \)

Let \( x - \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{y^2 - \left(\frac{3}{2}\right)^2 dy} \)

Apply the formula: \( \int \sqrt{x^2 - a^2 dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + C \)

Since \( \int \sqrt{y^2 - \left(\frac{3}{2}\right)^2 dy} \) is in the form \( \int \sqrt{x^2 - a^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{y^2 - \left(\frac{3}{2}\right)^2 dy} = \frac{y}{2}\sqrt{y^2 - \left(\frac{3}{2}\right)^2} - \frac{\left(\frac{3}{2}\right)^2}{2} \log |y + \sqrt{y^2 - \left(\frac{3}{2}\right)^2}| + C \)

\( = \frac{y}{2}\sqrt{y^2 - \frac{9}{4}} - \frac{9}{8} \log |y + \sqrt{y^2 - \frac{9}{4}}| + C \)

Since \( x - \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{x^2 - x - 2 dx} = \frac{1}{2}(2x - 1)\sqrt{x^2 - x - 2} - \frac{9}{8} \log \left|x - \frac{1}{2} + \sqrt{x^2 - x - 2}\right| + C \)

Now,
\( 3\int \sqrt{x^2 - x - 2 dx} = 3\left[\frac{1}{2}(2x - 1)\sqrt{x^2 - x - 2} - \frac{9}{8} \log \left|x - \frac{1}{2} + \sqrt{x^2 - x - 2}\right|\right] + C \)

\( = \frac{3}{2}(2x - 1)\sqrt{x^2 - x - 2} - \frac{27}{8} \log \left|x - \frac{1}{2} + \sqrt{x^2 - x - 2}\right| + C \)

Therefore, \( \int (4x + 1)\sqrt{x^2 - x - 2 dx} = \frac{4}{3}(x^2 - x - 2)^{\frac{3}{2}} + \frac{3}{2}(2x - 1)\sqrt{x^2 - x - 2} - \frac{27}{8} \log \left|x - \frac{1}{2} + \sqrt{x^2 - x - 2}\right| + C \)
In simple words: Break the linear coefficient into a derivative part and a constant. The derivative part integrates via substitution; the constant part requires completing the square and applying the difference formula.

Exam Tip: When decomposing \( 4x + 1 \) relative to derivative \( 2x - 1 \), write \( 4x + 1 = 2(2x - 1) + 3 \) and verify by expanding. The constant is found by \( 4x + 1 - 2(2x - 1) = 4x + 1 - 4x + 2 = 3 \).

Question 21. Evaluate the following integrals: \( \int (x + 1)\sqrt{2x^2 + 3dx} \)
Answer: To find: \( \int (x + 1)\sqrt{2x^2 + 3 dx} \)

Rewrite the integral as:

\( \int (x + 1)\sqrt{2x^2 + 3 dx} = \int [x\sqrt{2x^2 + 3} + \sqrt{2x^2 + 3}]dx \)

\( = \int x\sqrt{2x^2 + 3 dx} + \int \sqrt{2x^2 + 3 dx} \)

For the first integral, let \( 2x^2 + 3 = u \Rightarrow dx = \frac{du}{4x} \)

Thus, \( \int x\sqrt{2x^2 + 3 dx} = \frac{1}{4}\int \sqrt{u} du = \frac{1}{4} \cdot \frac{2}{3}u^{\frac{3}{2}} = \frac{1}{6}(2x^2 + 3)^{\frac{3}{2}} \)

For the second integral, rewrite \( \int \sqrt{2x^2 + 3 dx} \) as \( \int \sqrt{(\sqrt{2}x)^2 + (\sqrt{3})^2 dx} \)

Apply the formula: \( \int \sqrt{x^2 + a^2 dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \)

Since \( \int \sqrt{(\sqrt{2}x)^2 + (\sqrt{3})^2 dx} \) is of the form \( \int \sqrt{x^2 + a^2 dx} \),

\( \Rightarrow \int \sqrt{2x^2 + 3 dx} = \frac{\sqrt{2}x}{2}\sqrt{(\sqrt{2}x)^2 + (\sqrt{3})^2} + \frac{(\sqrt{3})^2}{2} \log |\sqrt{2}x + \sqrt{(\sqrt{2}x)^2 + (\sqrt{3})^2}| + C \)

\( = \frac{\sqrt{2}x}{2}\sqrt{2x^2 + 3} + \frac{3}{2\sqrt{2}} \log |\sqrt{2}x + \sqrt{2x^2 + 3}| + C \)

Therefore, \( \int (x + 1)\sqrt{2x^2 + 3 dx} = \frac{1}{6}(2x^2 + 3)^{\frac{3}{2}} + \frac{\sqrt{2}x}{2}\sqrt{2x^2 + 3} + \frac{3}{2\sqrt{2}} \log |\sqrt{2}x + \sqrt{2x^2 + 3}| + C \)
In simple words: Split the expression into two integrals. For the \( x \) part, use substitution with the radicand. For the constant part, extract the coefficient and apply the standard sum formula.

Exam Tip: When integrating \( x\sqrt{2x^2 + 3} \), the substitution \( u = 2x^2 + 3 \) immediately gives \( du = 4x dx \), so \( x dx = \frac{du}{4} \). This turns the integral into a simple power integral.

Question 22. Evaluate the following integrals: \( \int x\sqrt{1 + x - x^2 dx} \)
Answer: To find: \( \int x\sqrt{1 + x - x^2 dx} \)

Rewrite \( x \) as \( \frac{1}{2} - \frac{1}{2}(1 - 2x) \) and split:

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

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

For the first integral, let \( -x^2 + x + 1 = u \Rightarrow dx = \frac{du}{2x-1} \)

Thus, \( \frac{1}{2}\int (2x - 1)\sqrt{1 + x - x^2 dx} = -\frac{1}{2}\int \sqrt{u} du = -\frac{1}{2} \cdot \frac{2}{3}u^{\frac{3}{2}} = -\frac{1}{3}(-x^2 + x + 1)^{\frac{3}{2}} \)

For the second integral, complete the square:

\( \int \sqrt{1 + x - x^2 dx} = \int \sqrt{1 + x - x^2 - \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 dx} = \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2 dx} \)

Let \( x - \frac{1}{2} = y \Rightarrow dx = dy \)

Rewrite as \( \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2 dy} \)

Apply the formula: \( \int \sqrt{a^2 - x^2 dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\frac{x}{a} + C \)

Since \( \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2 dy} \) is in the form \( \int \sqrt{a^2 - x^2 dx} \) with a change in variable,

Hence, \( \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2 dy} = \frac{1}{2}y\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2} + \frac{\left(\frac{\sqrt{5}}{2}\right)^2}{2} \sin^{-1}\frac{y}{\frac{\sqrt{5}}{2}} + C \)

\( = \frac{y}{2}\sqrt{\frac{5}{4} - y^2} + \frac{5}{8} \sin^{-1}\frac{2y}{\sqrt{5}} + C \)

Since \( x - \frac{1}{2} = y \) and \( dx = dy \),

\( \Rightarrow \int \sqrt{1 + x - x^2 dx} = \frac{1}{2}\left(x - \frac{1}{2}\right)\sqrt{1 + x - x^2} + \frac{5}{8} \sin^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C \)

Now,
\( \frac{1}{2}\int \sqrt{1 + x - x^2 dx} = \frac{1}{4}\left(2x - 1\right)\sqrt{1 + x - x^2} + \frac{5}{16} \sin^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C \)

Therefore, \( \int x\sqrt{1 + x - x^2 dx} = -\frac{1}{3}(-x^2 + x + 1)^{\frac{3}{2}} + \frac{1}{4}(2x - 1)\sqrt{1 + x - x^2} + \frac{5}{16} \sin^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C \)
In simple words: Decompose the linear term \( x \) into parts that match the derivative of the radicand and a constant remainder. Apply substitution and the power rule for the first part. Complete the square and use the inverse sine formula for the second part.

Exam Tip: When decomposing \( x \) relative to derivative \( 2x - 1 \), solve algebraically: \( x = a(2x - 1) + b \Rightarrow x = 2ax - a + b \), so \( 2a = 1 \) (giving \( a = \frac{1}{2} \)) and \( -a + b = 0 \) (giving \( b = \frac{1}{2} \)).

Question 23. Evaluate the following integrals:
\( \int(2x - 5)\sqrt{2 + 3x - x^2}\,dx \)
Answer: To evaluate this integral, express \( 2x - 5 \) as \( (2x - 3) - 2 \) and split the integral into two parts.

Therefore,
\[ \int(2x - 5)\sqrt{2 + 3x - x^2}\,dx = \int(2x - 3)\sqrt{-x^2 + 3x + 2}\,dx - 2\int\sqrt{-x^2 + 3x + 2}\,dx \]

For the first integral, set \( -x^2 + 3x + 2 = u \), so \( dx = \frac{du}{3 - 2x} \).

Thus, \( \int(2x - 3)\sqrt{-x^2 + 3x + 2}\,dx \) becomes \( -\int\sqrt{u}\,du = -\frac{2}{3}u^{\frac{3}{2}} = -\frac{2}{3}(-x^2 + 3x + 2)^{\frac{3}{2}} \)

For the second integral, rewrite \( -x^2 + 3x + 2 \) in standard form:
\[ -x^2 + 3x + 2 = -\left(x - \frac{3}{2}\right)^2 + \frac{17}{4} \]

Set \( x - \frac{3}{2} = y \), so \( dx = dy \).

Then \( \int\sqrt{-x^2 + 3x + 2}\,dx \) becomes \( \int\sqrt{\left(\frac{\sqrt{17}}{2}\right)^2 - y^2}\,dy \)

Using the formula \( \int\sqrt{a^2 - x^2}\,dx = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \):

\[ \int\sqrt{\left(\frac{\sqrt{17}}{2}\right)^2 - y^2}\,dy = \frac{1}{2}y\sqrt{\left(\frac{\sqrt{17}}{2}\right)^2 - y^2} + \frac{17}{8}\sin^{-1}\frac{2y}{\sqrt{17}} + C \]

Substituting back \( y = x - \frac{3}{2} \):
\[ \int\sqrt{-x^2 + 3x + 2}\,dx = \frac{1}{4}(2x - 3)\sqrt{-x^2 + 3x + 2} + \frac{17}{8}\sin^{-1}\left(\frac{2x - 3}{\sqrt{17}}\right) + C \]

Combining both results:
\[ \int(2x - 5)\sqrt{2 + 3x - x^2}\,dx = -\frac{2}{3}(-x^2 + 3x + 2)^{\frac{3}{2}} - \frac{1}{2}(2x - 3)\sqrt{-x^2 + 3x + 2} - \frac{17}{4}\sin^{-1}\left(\frac{2x - 3}{\sqrt{17}}\right) + C \]

Exam Tip: Always decompose the linear term to match the derivative of the expression under the square root - this is essential for simplifying one part of the split integral. For the remaining square root integral, complete the square to get the standard inverse sine form.

Question 24. Evaluate the following integrals:
\( \int(6x + 5)\sqrt{6 + x - 2x^2}\,dx \)
Answer: To evaluate this integral, express \( 6x + 5 \) as \( \frac{13}{2} - \frac{3}{2}(1 - 4x) \) and divide into two separate integrals.

Therefore,
\[ \int(6x + 5)\sqrt{6 + x - 2x^2}\,dx = \int(4x - 1)\sqrt{-2x^2 + x + 6}\,dx + \int\sqrt{-2x^2 + x + 6}\,dx \]

For the first integral, set \( -2x^2 + x + 6 = u \), so \( dx = \frac{du}{1 - 4x} \).

Thus, \( \int(4x - 1)\sqrt{-2x^2 + x + 6}\,dx \) becomes \( -\int\sqrt{u}\,du = -\frac{2}{3}u^{\frac{3}{2}} = -\frac{2}{3}(-2x^2 + x + 6)^{\frac{3}{2}} \)

For the second integral, rewrite \( -2x^2 + x + 6 \) in standard form:
\[ -2x^2 + x + 6 = -2\left(x - \frac{1}{4}\right)^2 + \frac{49}{8} \]

Set \( x - \frac{1}{4} = y \), so \( dx = dy \).

Then \( \int\sqrt{-2x^2 + x + 6}\,dx \) becomes \( \int\sqrt{\left(\frac{7}{2\sqrt{2}}\right)^2 - y^2}\,dy \)

Using the standard formula:
\[ \int\sqrt{\left(\frac{7}{2\sqrt{2}}\right)^2 - y^2}\,dy = \frac{1}{2}y\sqrt{\left(\frac{7}{2\sqrt{2}}\right)^2 - y^2} + \frac{49}{16}\sin^{-1}\frac{2\sqrt{2}y}{7} + C \]

Substituting back \( y = x - \frac{1}{4} \):
\[ \int\sqrt{-2x^2 + x + 6}\,dx = \frac{1}{4}(4x - 1)\sqrt{-2x^2 + x + 6} + \frac{49}{16}\sin^{-1}\left(\frac{4x - 1}{7}\right) + C \]

Combining both results:
\[ \int(6x + 5)\sqrt{6 + x - 2x^2}\,dx = -\frac{2}{3}(-2x^2 + x + 6)^{\frac{3}{2}} + \frac{1}{4}(4x - 1)\sqrt{-2x^2 + x + 6} + \frac{49}{16}\sin^{-1}\left(\frac{4x - 1}{7}\right) + C \]

Exam Tip: When the coefficient of \( x^2 \) is not 1, factor it out before completing the square. This ensures the inverse sine argument stays within the valid \( [-1, 1] \) range.

Question 25. Evaluate the following integrals:
\( \int(x + 1)\sqrt{1 - x - x^2}\,dx \)
Answer: To evaluate this integral, express \( x + 1 \) as \( \frac{1}{2} - \frac{1}{2}(2x + 1) \) and divide into two integrals.

Therefore,
\[ \int(x + 1)\sqrt{1 - x - x^2}\,dx = \frac{1}{2}\int(2x - 1)\sqrt{-x^2 - x + 1}\,dx + \frac{1}{2}\int\sqrt{-x^2 - x + 1}\,dx \]

For the first integral, set \( -x^2 - x + 1 = u \), so \( dx = \frac{du}{-2x - 1} \).

Thus, \( \int(2x - 1)\sqrt{-x^2 - x + 1}\,dx \) becomes \( -\int\sqrt{u}\,du = -\frac{2}{3}u^{\frac{3}{2}} = -\frac{2}{3}(-x^2 - x + 1)^{\frac{3}{2}} \)

For the second integral, rewrite \( -x^2 - x + 1 \) in standard form:
\[ -x^2 - x + 1 = -\left(x + \frac{1}{2}\right)^2 + \frac{5}{4} \]

Set \( x + \frac{1}{2} = y \), so \( dx = dy \).

Then \( \int\sqrt{-x^2 - x + 1}\,dx \) becomes \( \int\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2}\,dy \)

Using the standard formula:
\[ \int\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2}\,dy = \frac{1}{2}y\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - y^2} + \frac{5}{8}\sin^{-1}\frac{2y}{\sqrt{5}} + C \]

Substituting back \( y = x + \frac{1}{2} \):
\[ \int\sqrt{-x^2 - x + 1}\,dx = \frac{1}{2}(2x + 1)\sqrt{-x^2 - x + 1} + \frac{5}{8}\sin^{-1}\left(\frac{2x + 1}{\sqrt{5}}\right) + C \]

Combining both results:
\[ \int(x + 1)\sqrt{1 - x - x^2}\,dx = -\frac{1}{3}(-x^2 - x + 1)^{\frac{3}{2}} + \frac{1}{4}(2x + 1)\sqrt{-x^2 - x + 1} + \frac{5}{16}\sin^{-1}\left(\frac{2x + 1}{\sqrt{5}}\right) + C \]

Exam Tip: When splitting integrals with linear numerators, ensure the coefficients in your decomposition are chosen so that one part has a derivative matching the radicand's derivative exactly, making substitution straightforward.

Question 26. Evaluate the following integrals:
\( \int(x - 3)\sqrt{x^2 + 3x - 18}\,dx \)
Answer: To evaluate this integral, express \( x - 3 \) as \( \frac{1}{2}(2x + 3) - \frac{9}{2} \) and divide into two integrals.

Therefore,
\[ \int(x - 3)\sqrt{x^2 + 3x - 18}\,dx = \frac{1}{2}\int(2x + 3)\sqrt{x^2 + 3x - 18}\,dx - \frac{9}{2}\int\sqrt{x^2 + 3x - 18}\,dx \]

For the first integral, set \( x^2 + 3x - 18 = u \), so \( dx = \frac{du}{2x + 3} \).

Thus, \( \int(2x + 3)\sqrt{x^2 + 3x - 18}\,dx \) becomes \( \int\sqrt{u}\,du = \frac{2}{3}u^{\frac{3}{2}} = \frac{2}{3}(x^2 + 3x - 18)^{\frac{3}{2}} \)

For the second integral, rewrite \( x^2 + 3x - 18 \) in standard form:
\[ x^2 + 3x - 18 = \left(x + \frac{3}{2}\right)^2 - \frac{81}{4} \]

Set \( x + \frac{3}{2} = y \), so \( dx = dy \).

Then \( \int\sqrt{x^2 + 3x - 18}\,dx \) becomes \( \int\sqrt{y^2 - \left(\frac{9}{2}\right)^2}\,dy \)

Using the formula \( \int\sqrt{x^2 - a^2}\,dx = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2}\log|x + \sqrt{x^2 - a^2}| + C \):

\[ \int\sqrt{y^2 - \left(\frac{9}{2}\right)^2}\,dy = \frac{1}{2}y\sqrt{y^2 - \frac{81}{4}} - \frac{81}{8}\log\left|y + \sqrt{y^2 - \frac{81}{4}}\right| + C \]

Substituting back \( y = x + \frac{3}{2} \):
\[ \int\sqrt{x^2 + 3x - 18}\,dx = \frac{1}{4}(2x + 3)\sqrt{x^2 + 3x - 18} - \frac{81}{8}\log\left|x + \frac{3}{2} + \sqrt{x^2 + 3x - 18}\right| + C \]

Combining both results:
\[ \int(x - 3)\sqrt{x^2 + 3x - 18}\,dx = \frac{1}{3}(x^2 + 3x - 18)^{\frac{3}{2}} - \frac{1}{8}(2x + 3)\sqrt{x^2 + 3x - 18} + \frac{726}{16}\log\left|x + \frac{3}{2} + \sqrt{x^2 + 3x - 18}\right| + C \]

Exam Tip: For integrals with a positive \( x^2 \) term under the square root, the standard formula involves a logarithmic term rather than an inverse sine. Always identify whether the radicand is of the form \( a^2 - x^2 \) or \( x^2 - a^2 \) to select the correct antiderivative formula.