NCERT Solutions · Class 12 · Chapter 7: Integrals

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.5

Exercise 7.5: Integration by partial fractions. Make the fraction proper first (divide if needed). Then \(\dfrac{px + q}{(x - a)(x - b)} = \dfrac{A}{x - a} + \dfrac{B}{x - b}\), a repeated factor \((x - a)^2\) gives \(\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}\), and an irreducible quadratic gives \(\dfrac{Bx + C}{x^2 + bx + c}\). Find constants by putting convenient x values (cover-up) and comparing coefficients.

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Exercise 7.5 questions and solutions

Exercise 7.5, Question 1

\(\displaystyle\int \dfrac{x}{(x + 1)(x + 2)}\,dx\)

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\[\dfrac{x}{(x + 1)(x + 2)} = \dfrac{-1}{x + 1} + \dfrac{2}{x + 2}\] (cover-up at \(x = -1\) and \(-2\)).

Answer: \(\log\dfrac{(x + 2)^2}{|x + 1|} + C\)

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Exercise 7.5, Question 2

\(\displaystyle\int \dfrac{1}{x^2 - 9}\,dx\)

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\[\dfrac{1}{(x - 3)(x + 3)} = \dfrac16\left(\dfrac{1}{x - 3} - \dfrac{1}{x + 3}\right)\]

Answer: \[\tfrac16\log\left|\dfrac{x - 3}{x + 3}\right| + C\]

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Exercise 7.5, Question 3

\(\displaystyle\int \dfrac{3x - 1}{(x - 1)(x - 2)(x - 3)}\,dx\)

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Cover-up: \(A = \dfrac{2}{(-1)(-2)} = 1\), \(B = \dfrac{5}{(1)(-1)} = -5\), \(C = \dfrac{8}{(2)(1)} = 4\).

Answer: \[\log|x - 1| - 5\log|x - 2| + 4\log|x - 3| + C\]

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Exercise 7.5, Question 4

\(\displaystyle\int \dfrac{x}{(x - 1)(x - 2)(x - 3)}\,dx\)

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\(A = \dfrac{1}{(-1)(-2)} = \tfrac12\), \(B = \dfrac{2}{(1)(-1)} = -2\), \(C = \dfrac{3}{(2)(1)} = \tfrac32\).

Answer: \[\tfrac12\log|x - 1| - 2\log|x - 2| + \tfrac32\log|x - 3| + C\]

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Exercise 7.5, Question 5

\(\displaystyle\int \dfrac{2x}{x^2 + 3x + 2}\,dx\)

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\[\dfrac{2x}{(x + 1)(x + 2)} = \dfrac{-2}{x + 1} + \dfrac{4}{x + 2}\]

Answer: \(4\log|x + 2| - 2\log|x + 1| + C\)

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Exercise 7.5, Question 6

\(\displaystyle\int \dfrac{1 - x^2}{x(1 - 2x)}\,dx\)

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Improper: \[\dfrac{x^2 - 1}{2x^2 - x} = \dfrac12 + \dfrac{\frac{x}{2} - 1}{x(2x - 1)}\]
\[\dfrac{\frac{x}{2} - 1}{x(2x - 1)} = \dfrac1x - \dfrac{3/2}{2x - 1}\] (cover-up at \(x = 0\) and \(\tfrac12\)).

Answer: \[\dfrac{x}{2} + \log|x| - \tfrac34\log|2x - 1| + C\]

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Exercise 7.5, Question 7

\(\displaystyle\int \dfrac{x}{(x^2 + 1)(x - 1)}\,dx\)

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\[\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 1}\]: \(A = \tfrac12\); comparing \(x^2\) and constant terms: \(A + B = 0\), \(A - C = 0\), so \(B = -\tfrac12\), \(C = \tfrac12\).

Answer: \[\tfrac12\log|x - 1| - \tfrac14\log(x^2 + 1) + \tfrac12\tan^{-1}x + C\]

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Exercise 7.5, Question 8

\(\displaystyle\int \dfrac{x}{(x - 1)^2(x + 2)}\,dx\)

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\[\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2} + \dfrac{C}{x + 2}\]: \(C = \dfrac{-2}{9}\), \(B = \dfrac13\), and \(A + C = 0\) gives \(A = \tfrac29\).

Answer: \[\tfrac29\log\left|\dfrac{x - 1}{x + 2}\right| - \dfrac{1}{3(x - 1)} + C\]

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Exercise 7.5, Question 9

\(\displaystyle\int \dfrac{3x + 5}{x^3 - x^2 - x + 1}\,dx\)

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\(x^3 - x^2 - x + 1 = (x - 1)^2(x + 1)\).
\(C = \dfrac{2}{4} = \tfrac12\) (for \(x + 1\)), \(B = \dfrac{8}{2} = 4\) (for \((x - 1)^2\)), \(A = -C = -\tfrac12\).

Answer: \[\tfrac12\log\left|\dfrac{x + 1}{x - 1}\right| - \dfrac{4}{x - 1} + C\]

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Exercise 7.5, Question 10

\(\displaystyle\int \dfrac{2x - 3}{(x^2 - 1)(2x + 3)}\,dx\)

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\[\dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{2x + 3}\]: \(A = \dfrac{-1}{(2)(5)} = -\tfrac{1}{10}\), \(B = \dfrac{-5}{(-2)(1)} = \tfrac52\), \[\begin{aligned}C &= \dfrac{-6}{\left(-\frac52\right)\left(-\frac12\right)} \\ &= -\tfrac{24}{5}\end{aligned}\]
\[\int\dfrac{-24/5}{2x + 3}\,dx = -\tfrac{12}{5}\log|2x + 3|\]

Answer: \[\tfrac52\log|x + 1| - \tfrac{1}{10}\log|x - 1| - \tfrac{12}{5}\log|2x + 3| + C\]

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Exercise 7.5, Question 11

\(\displaystyle\int \dfrac{5x}{(x + 1)(x^2 - 4)}\,dx\)

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Factors \(x + 1, x - 2, x + 2\): \(A = \dfrac{-5}{(-3)(1)} = \tfrac53\), \(B = \dfrac{10}{(3)(4)} = \tfrac56\), \(C = \dfrac{-10}{(-1)(-4)} = -\tfrac52\).

Answer: \[\tfrac53\log|x + 1| + \tfrac56\log|x - 2| - \tfrac52\log|x + 2| + C\]

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Exercise 7.5, Question 12

\(\displaystyle\int \dfrac{x^3 + x + 1}{x^2 - 1}\,dx\)

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Divide: \(x + \dfrac{2x + 1}{x^2 - 1}\).
\[\dfrac{2x + 1}{(x - 1)(x + 1)} = \dfrac{3/2}{x - 1} + \dfrac{1/2}{x + 1}\]

Answer: \[\dfrac{x^2}{2} + \tfrac32\log|x - 1| + \tfrac12\log|x + 1| + C\]

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Exercise 7.5, Question 13

\(\displaystyle\int \dfrac{2}{(1 - x)(1 + x^2)}\,dx\)

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\[\dfrac{A}{1 - x} + \dfrac{Bx + C}{1 + x^2}\]: \(A = 1\); \(x^2\): \(A - B = 0\); constants: \(A + C = 2\). So \(B = 1\), \(C = 1\).

Answer: \[-\log|1 - x| + \tfrac12\log(1 + x^2) + \tan^{-1}x + C\]

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Exercise 7.5, Question 14

\(\displaystyle\int \dfrac{3x - 1}{(x + 2)^2}\,dx\)

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\(3x - 1 = 3(x + 2) - 7\): \(\dfrac{3}{x + 2} - \dfrac{7}{(x + 2)^2}\).

Answer: \(3\log|x + 2| + \dfrac{7}{x + 2} + C\)

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Exercise 7.5, Question 15

\(\displaystyle\int \dfrac{1}{x^4 - 1}\,dx\)

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\[\dfrac{1}{(x^2 - 1)(x^2 + 1)} = \dfrac12\left[\dfrac{1}{x^2 - 1} - \dfrac{1}{x^2 + 1}\right]\]

Answer: \[\tfrac14\log\left|\dfrac{x - 1}{x + 1}\right| - \tfrac12\tan^{-1}x + C\]

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Exercise 7.5, Question 16

\(\displaystyle\int\dfrac{dx}{x(x^n + 1)}\) (multiply top and bottom by \(x^{n-1}\), put \(x^n = t\))

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\(\dfrac{x^{n-1}}{x^n(x^n + 1)}\); \(t = x^n\), \(dt = nx^{n-1}\,dx\): \[\dfrac1n\int\dfrac{dt}{t(t + 1)} = \dfrac1n\int\left(\dfrac1t - \dfrac{1}{t + 1}\right)dt\]

Answer: \[\dfrac1n\log\left|\dfrac{x^n}{x^n + 1}\right| + C\]

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Exercise 7.5, Question 17

\(\displaystyle\int \dfrac{\cos x}{(1 - \sin x)(2 - \sin x)}\,dx\)

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\(t = \sin x\): \[\begin{aligned}\int\dfrac{dt}{(1 - t)(2 - t)} &= \int\left(\dfrac{1}{1 - t} - \dfrac{1}{2 - t}\right)dt \\ &= -\log|1 - t| + \log|2 - t|\end{aligned}\]

Answer: \[\log\left|\dfrac{2 - \sin x}{1 - \sin x}\right| + C\]

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Exercise 7.5, Question 18

\(\displaystyle\int \dfrac{(x^2 + 1)(x^2 + 2)}{(x^2 + 3)(x^2 + 4)}\,dx\)

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With \(y = x^2\): \[\begin{aligned}\dfrac{(y + 1)(y + 2)}{(y + 3)(y + 4)} &= 1 - \dfrac{4y + 10}{(y + 3)(y + 4)} \\ &= 1 + \dfrac{2}{y + 3} - \dfrac{6}{y + 4}\end{aligned}\]

Answer: \[x + \dfrac{2}{\sqrt3}\tan^{-1}\dfrac{x}{\sqrt3} - 3\tan^{-1}\dfrac{x}{2} + C\]

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Exercise 7.5, Question 19

\(\displaystyle\int \dfrac{2x}{(x^2 + 1)(x^2 + 3)}\,dx\)

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\(t = x^2\): \[\int\dfrac{dt}{(t + 1)(t + 3)} = \tfrac12\int\left(\dfrac{1}{t + 1} - \dfrac{1}{t + 3}\right)dt\]

Answer: \(\tfrac12\log\dfrac{x^2 + 1}{x^2 + 3} + C\)

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Exercise 7.5, Question 20

\(\displaystyle\int \dfrac{1}{x(x^4 - 1)}\,dx\)

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Multiply by \(\dfrac{x^3}{x^3}\); \(t = x^4\): \[\tfrac14\int\dfrac{dt}{t(t - 1)} = \tfrac14\int\left(\dfrac{1}{t - 1} - \dfrac1t\right)dt\]

Answer: \[\tfrac14\log\left|\dfrac{x^4 - 1}{x^4}\right| + C\]

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Exercise 7.5, Question 21

\(\displaystyle\int \dfrac{1}{e^x - 1}\,dx\)

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\(t = e^x\), \(dx = \dfrac{dt}{t}\): \[\int\dfrac{dt}{t(t - 1)} = \log\left|\dfrac{t - 1}{t}\right|\]

Answer: \[\log\left|\dfrac{e^x - 1}{e^x}\right| + C\]

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Exercise 7.5, Question 22

\(\displaystyle\int\dfrac{x\,dx}{(x - 1)(x - 2)}\) equals: (A) \(\log\left|\dfrac{(x - 1)^2}{x - 2}\right| + C\) (B) \(\log\left|\dfrac{(x - 2)^2}{x - 1}\right| + C\) (C) \(\log\left|\left(\dfrac{x - 1}{x - 2}\right)^2\right| + C\) (D) \(\log|(x - 1)(x - 2)| + C\)

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\[\dfrac{x}{(x - 1)(x - 2)} = \dfrac{-1}{x - 1} + \dfrac{2}{x - 2}\]

Answer: (B)

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Exercise 7.5, Question 23

\(\displaystyle\int\dfrac{dx}{x(x^2 + 1)}\) equals: (A) \(\log|x| - \tfrac12\log(x^2 + 1) + C\) (B) \(\log|x| + \tfrac12\log(x^2 + 1) + C\) (C) \(-\log|x| + \tfrac12\log(x^2 + 1) + C\) (D) \(\tfrac12\log|x| + \log(x^2 + 1) + C\)

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\[\dfrac{1}{x(x^2 + 1)} = \dfrac1x - \dfrac{x}{x^2 + 1}\]

Answer: (A)

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