Maharashtra Board Class 12 Maths Part 1 Chapter 5 Integration 5.6 Solutions

Evaluate:

Question 1. \(\int \frac{2x+1}{(x+1)(x-2)} dx\)
Answer:
Solution:
Let \(I = \int \frac{2x+1}{(x+1)(x-2)} dx\)
Let \(\frac{2x+1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}\)
\( \implies \) \(2x + 1 = A(x - 2) + B(x + 1)\)
Put \(x + 1 = 0\), i.e. \(x = -1\), we get
\(2(-1) + 1 = A(-3) + B(0)\)
\(-1 = -3A\)
\(\therefore A = \frac{1}{3}\)
Put \(x - 2 = 0\), i.e. \(x = 2\), we get
\(2(2) + 1 = A(0) + B(3)\)
\(5 = 3B\)
\(\therefore B = \frac{5}{3}\)
\(\therefore \frac{2x+1}{(x+1)(x-2)} = \frac{(1/3)}{x+1} + \frac{(5/3)}{x-2}\)
\(\therefore I = \int \left[ \frac{(1/3)}{x+1} + \frac{(5/3)}{x-2} \right] dx\)
\( = \frac{1}{3} \int \frac{1}{x+1} dx + \frac{5}{3} \int \frac{1}{x-2} dx\)
\( = \frac{1}{3} \log|x + 1| + \frac{5}{3} \log|x - 2| + c\).
In simple words: This problem involves integrating a rational function by decomposing it into partial fractions. First, we find the constants A and B using the substitution method, then integrate each simpler fraction, resulting in logarithmic terms.

🎯 Exam Tip: Mastering partial fraction decomposition is crucial for solving such integration problems. Always double-check the values of constants A, B, and C as errors here propagate through the entire solution.

Question 2. \(\int \frac{2x+1}{x(x-1)(x-4)} dx\)
Answer:
Solution:
Let \(I = \int \frac{2x+1}{x(x-1)(x-4)} dx\)
Let \(\frac{2x+1}{x(x-1)(x-4)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-4}\)
\( \implies \) \(2x + 1 = A(x - 1)(x - 4) + Bx(x - 4) + Cx(x - 1)\)
Put \(x = 0\), we get
\(2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1)\)
\(1 = 4A\)
\(\therefore A = \frac{1}{4}\)
Put \(x - 1 = 0\), i.e. \(x = 1\), we get
\(2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0)\)
\(3 = -3B\)
\(\therefore B = -1\)
Put \(x - 4 = 0\), i.e \(x = 4\), we get
\(2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3)\)
\(9 = 12C\)
\(\therefore C = \frac{3}{4}\)
\(\therefore \frac{2x+1}{x(x-1)(x-4)} = \frac{(1/4)}{x} + \frac{(-1)}{x-1} + \frac{(3/4)}{x-4}\)
\(\therefore I = \int \left[ \frac{(1/4)}{x} - \frac{1}{x-1} + \frac{(3/4)}{x-4} \right] dx\)
\( = \frac{1}{4} \int \frac{1}{x} dx - \int \frac{1}{x-1} dx + \frac{3}{4} \int \frac{1}{x-4} dx\)
\( = \frac{1}{4} \log|x| - \log|x-1| + \frac{3}{4} \log|x-4] + c\).
In simple words: We integrate this rational expression by breaking it down into partial fractions with three distinct linear denominators. After finding the coefficients A, B, and C, we integrate each term, resulting in a sum of logarithmic functions.

🎯 Exam Tip: When dealing with multiple linear factors in the denominator, the cover-up method (or Heaviside's method) is often the quickest way to find the partial fraction coefficients. Present each step clearly to earn full marks.

Question 3. \(\int \frac{x^2+x-1}{x^2+x-6} dx\)
Answer:
Solution:
Let \(I = \int \frac{x^2+x-1}{x^2+x-6} dx\)
\( = \int \frac{(x^2+x-6)+5}{x^2+x-6} dx\)
\( = \int \left[ 1 + \frac{5}{x^2+x-6} \right] dx\)
\( = \int 1 dx + 5 \int \frac{1}{x^2+x-6} dx\)
Let \(\frac{1}{x^2+x-6} = \frac{1}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}\)
\( \implies \) \(1 = A(x - 2) + B(x + 3)\)
Put \(x + 3 = 0\), i.e. \(x = -3\), we get
\(1 = A(-5) + B(0)\)
\(\therefore A = -\frac{1}{5}\)
Put \(x - 2 = 0\), i.e. \(x = 2\), we get
\(1 = A(0) + B(5)\)
\(\therefore B = \frac{1}{5}\)
\(\therefore \frac{1}{x^2+x-6} = \frac{(-1/5)}{x+3} + \frac{(1/5)}{x-2}\)
\(\therefore I = \int 1 dx + 5 \int \left[ \frac{(-1/5)}{x+3} + \frac{(1/5)}{x-2} \right] dx\)
\( = \int 1 dx - \int \frac{1}{x+3} dx + \int \frac{1}{x-2} dx\)
\( = x - \log|x + 3| + \log|x - 2| + c\).
In simple words: When the degree of the numerator is equal to the degree of the denominator, we first perform polynomial long division (or algebraic manipulation) to separate a constant term. Then, the remaining rational function is integrated using partial fraction decomposition.

🎯 Exam Tip: Always check the degrees of the numerator and denominator. If the numerator's degree is greater than or equal to the denominator's, perform polynomial division before applying partial fractions.

Question 4. \(\int \frac{x}{(x-1)^2(x+2)} dx\)
Answer:
Solution:
Let \(I = \int \frac{x}{(x-1)^2(x+2)} dx\)
Let \(\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}\)
\( \implies \) \(x = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2\)
Put \(x - 1 = 0\), i.e. \(x = 1\), we get
\(1 = A(0)(3) + B(3) + C(0)\)
\(1 = 3B\)
\(\therefore B = \frac{1}{3}\)
Put \(x + 2 = 0\), i.e. \(x = -2\), we get
\(-2 = A(-3)(0) + B(0) + C(-3)^2\)
\(-2 = 9C\)
\(\therefore C = -\frac{2}{9}\)
Put \(x = 0\), we get (this is to find A, using a simple value)
\(0 = A(-1)(2) + B(2) + C(1)\)
\(0 = -2A + 2B + C\)
Substitute \(B = \frac{1}{3}\) and \(C = -\frac{2}{9}\)
\(0 = -2A + 2(\frac{1}{3}) + (-\frac{2}{9})\)
\(0 = -2A + \frac{2}{3} - \frac{2}{9}\)
\(2A = \frac{2}{3} - \frac{2}{9} = \frac{6-2}{9} = \frac{4}{9}\)
\(\therefore A = \frac{2}{9}\)
\(\therefore \frac{x}{(x-1)^2(x+2)} = \frac{(2/9)}{x-1} + \frac{(1/3)}{(x-1)^2} + \frac{(-2/9)}{x+2}\)
\(\therefore I = \int \left[ \frac{(2/9)}{x-1} + \frac{(1/3)}{(x-1)^2} + \frac{(-2/9)}{x+2} \right] dx\)
\( = \frac{2}{9} \int \frac{1}{x-1} dx + \frac{1}{3} \int (x-1)^{-2} dx - \frac{2}{9} \int \frac{1}{x+2} dx\)
\( = \frac{2}{9} \log|x-1| + \frac{1}{3} \frac{(x-1)^{-1}}{-1} - \frac{2}{9} \log|x + 2| + c\)
\( = \frac{2}{9} \log|x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log|x+2| + c\).
In simple words: This integral involves a repeated linear factor in the denominator. We set up the partial fractions with distinct linear and squared terms. After finding the coefficients by substituting convenient x-values and solving a system of equations, we integrate each term, which includes logarithmic and power rule integrations.

🎯 Exam Tip: For repeated factors like \((x-1)^2\), remember to include both \(\frac{A}{x-1}\) and \(\frac{B}{(x-1)^2}\) in the partial fraction decomposition. Don't forget the negative sign when integrating \((x-1)^{-2}\).

Question 5. \(\int \frac{3x-2}{(x+1)^2(x+3)} dx\)
Answer:
Solution:
Let \(I = \int \frac{3x-2}{(x+1)^2(x+3)} dx\)
Let \(\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}\)
\( \implies \) \(3x - 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)^2\)
Put \(x + 1 = 0\), i.e. \(x = -1\), we get
\(3(-1) - 2 = A(0)(2) + B(2) + C(0)\)
\(-5 = 2B\)
\(\therefore B = -\frac{5}{2}\)
Put \(x + 3 = 0\), i.e. \(x = -3\), we get
\(3(-3) - 2 = A(-2)(0) + B(0) + C(-2)^2\)
\(-11 = 4C\)
\(\therefore C = -\frac{11}{4}\)
Put \(x = 0\), we get
\(3(0) - 2 = A(1)(3) + B(3) + C(1)\)
\(-2 = 3A + 3B + C\)
Substitute \(B = -\frac{5}{2}\) and \(C = -\frac{11}{4}\)
\(-2 = 3A + 3(-\frac{5}{2}) + (-\frac{11}{4})\)
\(-2 = 3A - \frac{15}{2} - \frac{11}{4}\)
\(3A = -2 + \frac{15}{2} + \frac{11}{4}\)
\(3A = \frac{-8 + 30 + 11}{4} = \frac{33}{4}\)
\(\therefore A = \frac{11}{4}\)
\(\therefore \frac{3x-2}{(x+1)^2(x+3)} = \frac{(11/4)}{x+1} + \frac{(-5/2)}{(x+1)^2} + \frac{(-11/4)}{x+3}\)
\(\therefore I = \int \left[ \frac{(11/4)}{x+1} - \frac{(5/2)}{(x+1)^2} - \frac{(11/4)}{x+3} \right] dx\)
\( = \frac{11}{4} \int \frac{1}{x+1} dx - \frac{5}{2} \int (x+1)^{-2} dx - \frac{11}{4} \int \frac{1}{x+3} dx\)
\( = \frac{11}{4} \log|x+1| - \frac{5}{2} \frac{(x+1)^{-1}}{-1} - \frac{11}{4} \log|x+3| + c\)
\( = \frac{11}{4} \log\left|\frac{x+1}{x+3}\right| + \frac{5}{2(x+1)} + c\).
In simple words: This problem involves integrating a rational function with a repeated linear factor and another distinct linear factor in the denominator. We use partial fraction decomposition, solve for the constants, and then integrate each resulting term, combining logarithmic expressions where possible.

🎯 Exam Tip: When calculating constants for partial fractions, strategically choosing x-values that make terms zero simplifies the process. For coefficients that don't cancel, use an easy x-value like 0 or -1 and solve the resulting equation.

Question 6. \(\int \frac{1}{x(x^5+1)} dx\)
Answer:
Solution:
Let \(I = \int \frac{1}{x(x^5+1)} dx\)
Multiply numerator and denominator by \(x^4\):
\( = \int \frac{x^4}{x^5(x^5+1)} dx\)
Put \(x^5 = t\). Then \(5x^4 dx = dt\)
\(\therefore x^4 dx = \frac{dt}{5}\)
\(\therefore I = \int \frac{1}{t(t+1)} \frac{dt}{5}\)
\( = \frac{1}{5} \int \frac{1}{t(t+1)} dt\)
We know that \(\frac{1}{t(t+1)} = \frac{t+1-t}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}\)
\( = \frac{1}{5} \int \left[ \frac{1}{t} - \frac{1}{t+1} \right] dt\)
\( = \frac{1}{5} [\int \frac{1}{t} dt - \int \frac{1}{t+1} dt]\)
\( = \frac{1}{5} [\log|t| - \log|t+1|] + c\)
\( = \frac{1}{5} \log\left|\frac{t}{t+1}\right| + c\)
Substitute back \(t = x^5\):
\( = \frac{1}{5} \log\left|\frac{x^5}{x^5+1}\right| + c\).
In simple words: This integral requires a clever substitution to simplify the expression. By multiplying the numerator and denominator by \(x^4\) and substituting \(t = x^5\), the integral transforms into a simpler form that can be solved using basic partial fraction identity and logarithmic integration.

🎯 Exam Tip: Look for opportunities to create a derivative of a complex term. Here, multiplying by \(x^4\) created \(x^5\) and \(x^4 dx\), which perfectly fits for a substitution where \(t = x^5\).

Question 7. \(\int \frac{1}{x(x^n+1)} dx\)
Answer:
Solution:
Let \(I = \int \frac{1}{x(x^n+1)} dx\)
Multiply numerator and denominator by \(x^{n-1}\):
\( = \int \frac{x^{n-1}}{x^n(x^n+1)} dx\)
Put \(x^n = t\). Then \(nx^{n-1} dx = dt\)
\(\therefore x^{n-1} dx = \frac{dt}{n}\)
\(\therefore I = \int \frac{1}{t(t+1)} \frac{dt}{n}\)
\( = \frac{1}{n} \int \frac{1}{t(t+1)} dt\)
We know that \(\frac{1}{t(t+1)} = \frac{t+1-t}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}\)
\( = \frac{1}{n} \int \left[ \frac{1}{t} - \frac{1}{t+1} \right] dt\)
\( = \frac{1}{n} [\int \frac{1}{t} dt - \int \frac{1}{t+1} dt]\)
\( = \frac{1}{n} [\log|t| - \log|t+1|] + c\)
\( = \frac{1}{n} \log\left|\frac{t}{t+1}\right| + c\)
Substitute back \(t = x^n\):
\( = \frac{1}{n} \log\left|\frac{x^n}{x^n+1}\right| + c\).
In simple words: This generalizes the previous problem. By multiplying the numerator and denominator by \(x^{n-1}\) and performing the substitution \(t = x^n\), the integral simplifies to a basic form involving the difference of two logarithmic functions, which is then expressed in terms of the original variable.

🎯 Exam Tip: Recognizing patterns from previous problems is a powerful strategy. Here, the solution method is analogous to Question 6, but with a general exponent 'n'. Be precise with variable substitutions and back-substitutions.

Question 8. \(\int \frac{5x^2+20x+6}{x^3+2x^2+x} dx\)
Answer:
Solution:
Let \(I = \int \frac{5x^2+20x+6}{x^3+2x^2+x} dx\)
First, factor the denominator: \(x^3+2x^2+x = x(x^2+2x+1) = x(x+1)^2\)
So, \(I = \int \frac{5x^2+20x+6}{x(x+1)^2} dx\)
Let \(\frac{5x^2+20x+6}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}\)
\( \implies \) \(5x^2+20x+6 = A(x+1)^2 + Bx(x+1) + Cx\)
Put \(x = 0\), we get
\(5(0)^2+20(0)+6 = A(0+1)^2 + B(0)(0+1) + C(0)\)
\(6 = A(1) + 0 + 0\)
\(\therefore A = 6\)
Put \(x + 1 = 0\), i.e. \(x = -1\), we get
\(5(-1)^2+20(-1)+6 = A(0)^2 + B(-1)(0) + C(-1)\)
\(5 - 20 + 6 = -C\)
\(-9 = -C\)
\(\therefore C = 9\)
To find B, substitute \(A=6\), \(C=9\) and any other value for \(x\), say \(x=1\):
\(5(1)^2+20(1)+6 = A(1+1)^2 + B(1)(1+1) + C(1)\)
\(5+20+6 = A(2)^2 + B(2) + C\)
\(31 = 4A + 2B + C\)
\(31 = 4(6) + 2B + 9\)
\(31 = 24 + 2B + 9\)
\(31 = 33 + 2B\)
\(2B = 31 - 33\)
\(2B = -2\)
\(\therefore B = -1\)
\(\therefore \frac{5x^2+20x+6}{x(x+1)^2} = \frac{6}{x} - \frac{1}{x+1} + \frac{9}{(x+1)^2}\)
\(\therefore I = \int \left[ \frac{6}{x} - \frac{1}{x+1} + \frac{9}{(x+1)^2} \right] dx\)
\( = 6 \int \frac{1}{x} dx - \int \frac{1}{x+1} dx + 9 \int (x+1)^{-2} dx\)
\( = 6 \log|x| - \log|x+1| + 9 \frac{(x+1)^{-1}}{-1} + c\)
\( = 6 \log|x| - \log|x+1| - \frac{9}{x+1} + c\).
In simple words: This integral involves a rational function where the denominator can be factored into a distinct linear term and a repeated linear term. We use partial fraction decomposition, solving for A, B, and C by strategic substitution, and then integrate each term, resulting in a combination of logarithmic and power rule terms.

🎯 Exam Tip: Always factor the denominator first. For repeated factors, ensure you include all terms (e.g., for \((x+1)^2\), include both \(\frac{B}{x+1}\) and \(\frac{C}{(x+1)^2}\)). Remember that \(\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}\) for \(n \ne -1\).