Exercise 7.5 Class 12 Maths Integrals

NCERT Solutions Class 12 Maths Chapter 7 Ex 7.5 | LearnCBSEHub

💡 Decomposition Rules

We decompose proper rational functions $\frac{P(x)}{Q(x)}$ based on the denominator type:

Linear Factors: $\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
Repeated Linear: $\frac{1}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$
Quadratic Factor: $\frac{1}{x^2+bx+c} \to \frac{Ax+B}{x^2+bx+c}$

Question 1

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

Step 1: Partial Fractions

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

Step 2: Integrate

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

Question 2

$\int \frac{1}{x^2-9} dx$

Use formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log|\frac{x-a}{x+a}|$.
Here $a=3$.
$= \frac{1}{6}\log|\frac{x-3}{x+3}| + C$.

Question 3

$\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx$

Step 1: Form Partial Fractions

$\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$
$3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
$x=1 \implies 2 = A(2) \implies A=1$.
$x=2 \implies 5 = B(-1) \implies B=-5$.
$x=3 \implies 8 = C(2) \implies C=4$.

Step 2: Integrate

$\log|x-1| – 5\log|x-2| + 4\log|x-3| + C$

Question 4

$\int \frac{x}{(x-1)(x-2)(x-3)} dx$

Step 1: Find A, B, C

$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
$x=1: 1 = 2A \implies A=1/2$.
$x=2: 2 = -B \implies B=-2$.
$x=3: 3 = 2C \implies C=3/2$.

$\frac{1}{2}\log|x-1| – 2\log|x-2| + \frac{3}{2}\log|x-3| + C$

Question 5

$\int \frac{2x}{x^2+3x+2} dx$

Denominator factors to $(x+1)(x+2)$.
$\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.
$x=-1 \implies -2 = A(1) \implies A=-2$.
$x=-2 \implies -4 = B(-1) \implies B=4$.
Ans: $4\log|x+2| – 2\log|x+1| + C$.

Question 6

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

Step 1: Improper Fraction

Degree of num (2) = Degree of den (2). Perform long division.
$\frac{1-x^2}{x-2x^2} = \frac{1}{2} + \frac{1-x/2}{x(1-2x)} = \frac{1}{2} + \frac{2-x}{2x(1-2x)}$.
Let $\frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}$.
$2-x = A(1-2x) + Bx$.
$x=0 \implies A=2$. $x=1/2 \implies 3/2 = B/2 \implies B=3$.

$\frac{x}{2} + \log|x| – \frac{3}{4}\log|1-2x| + C$

Question 7

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

Step 1: Setup

$\frac{x}{(x^2+1)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$
$x = A(x^2+1) + (Bx+C)(x-1)$.
$x=1 \implies 1 = 2A \implies A=1/2$.
Compare coeffs of $x^2$: $0 = A+B \implies B=-1/2$.
Compare constant term: $0 = A-C \implies C=1/2$.

Step 2: Integrate

$\int (\frac{1/2}{x-1} + \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2+1}) dx$.
$= \frac{1}{2}\log|x-1| – \frac{1}{4}\log|x^2+1| + \frac{1}{2}\tan^{-1}x + C$.

Question 8

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

Step 1: Setup

$\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.
$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$.
$x=1 \implies 1 = 3B \implies B=1/3$.
$x=-2 \implies -2 = 9C \implies C=-2/9$.
Coeff $x^2$: $0 = A+C \implies A=2/9$.

$\frac{2}{9}\log|x-1| – \frac{1}{3(x-1)} – \frac{2}{9}\log|x+2| + C$

Question 9

$\int \frac{3x+5}{x^3-x^2-x+1} dx$

Denominator: $x^2(x-1) – 1(x-1) = (x^2-1)(x-1) = (x+1)(x-1)^2$.
$\frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$.
$x=-1 \implies 2 = 4A \implies A=1/2$.
$x=1 \implies 8 = 2C \implies C=4$.
Coeff $x^2$: $0 = A+B \implies B=-1/2$.
Ans: $\frac{1}{2}\log|x+1| – \frac{1}{2}\log|x-1| – \frac{4}{x-1} + C$.

Question 10

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

$\frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}$.
$A = -1/10, B = 5/2, C = -24/5$.
Ans: $\frac{5}{2}\log|x+1| – \frac{1}{10}\log|x-1| – \frac{12}{5}\log|2x+3| + C$.

Question 11

$\int \frac{5x}{(x+1)(x^2-4)} dx$

$\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}$.
$A=5/3, B=5/6, C=-5/2$.
Ans: $\frac{5}{3}\log|x+1| + \frac{5}{6}\log|x-2| – \frac{5}{2}\log|x+2| + C$.

Question 12

$\int \frac{x^3+x+1}{x^2-1} dx$

Improper fraction. Division: $\frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1}$.
$\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$. $A=3/2, B=1/2$.
Ans: $\frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C$.

Question 13

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

$= \frac{A}{1-x} + \frac{Bx+C}{1+x^2}$.
$2 = A(1+x^2) + (Bx+C)(1-x)$.
$x=1 \implies A=1$. $x=0 \implies 2=A+C \implies C=1$.
Coeff $x^2$: $0 = A-B \implies B=1$.
Ans: $-\log|1-x| + \frac{1}{2}\log|1+x^2| + \tan^{-1}x + C$.

Question 14

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

$= \frac{A}{x+2} + \frac{B}{(x+2)^2}$.
$3x-1 = A(x+2) + B$.
$x=-2 \implies B=-7$. Compare x: $3=A$.
Ans: $3\log|x+2| + \frac{7}{x+2} + C$.

Question 15

$\int \frac{1}{x^4-1} dx$

$\frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2}[\frac{1}{x^2-1} – \frac{1}{x^2+1}]$.
$= \frac{1}{2} [\frac{1}{2}\log|\frac{x-1}{x+1}| – \tan^{-1}x] + C$.
$= \frac{1}{4}\log|\frac{x-1}{x+1}| – \frac{1}{2}\tan^{-1}x + C$.

Question 16

$\int \frac{1}{x(x^n+1)} dx$

Multiply num/den by $x^{n-1}$. $\int \frac{x^{n-1}}{x^n(x^n+1)} dx$.
Let $x^n = t \implies nx^{n-1} dx = dt$.
$\frac{1}{n} \int \frac{dt}{t(t+1)} = \frac{1}{n} \int (\frac{1}{t} – \frac{1}{t+1}) dt$.
$= \frac{1}{n} \log|\frac{x^n}{x^n+1}| + C$.

Question 17

$\int \frac{\cos x}{(1-\sin x)(2-\sin x)} dx$

Let $\sin x = t \implies \cos x dx = dt$.
$\int \frac{dt}{(1-t)(2-t)} = \int (\frac{1}{1-t} – \frac{1}{2-t}) dt$.
$= -\log|1-t| + \log|2-t| = \log|\frac{2-\sin x}{1-\sin x}| + C$.

Question 18

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

Let $x^2 = y$ (for partial fraction only).
$\frac{(y+1)(y+2)}{(y+3)(y+4)} = 1 – \frac{4y+10}{(y+3)(y+4)} = 1 – [\frac{-2}{y+3} + \frac{6}{y+4}]$.
Integrate w.r.t $x$: $\int (1 + \frac{2}{x^2+3} – \frac{6}{x^2+4}) dx$.
Ans: $x + \frac{2}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}}) – 3\tan^{-1}(\frac{x}{2}) + C$.

Question 19

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

Let $x^2 = t \implies 2x dx = dt$.
$\int \frac{dt}{(t+1)(t+3)} = \frac{1}{2} \int (\frac{1}{t+1} – \frac{1}{t+3}) dt$.
$= \frac{1}{2}\log|\frac{x^2+1}{x^2+3}| + C$.

Question 20

$\int \frac{1}{x(x^4-1)} dx$

Similar to Q16 with $n=4$ and minus sign.
Multiply by $x^3$: $\int \frac{x^3}{x^4(x^4-1)} dx$. Let $x^4=t$.
$\frac{1}{4} \int \frac{dt}{t(t-1)} = \frac{1}{4}\log|\frac{t-1}{t}| = \frac{1}{4}\log|\frac{x^4-1}{x^4}| + C$.

Question 21

$\int \frac{1}{e^x-1} dx$

Let $e^x = t \implies e^x dx = dt \implies dx = dt/t$.
$\int \frac{dt}{t(t-1)} = \log|\frac{t-1}{t}| = \log|\frac{e^x-1}{e^x}| + C$.

Question 22

$\int \frac{x}{(x-1)(x-2)} dx$

$= \frac{-1}{x-1} + \frac{2}{x-2}$.
Ans: $-\log|x-1| + 2\log|x-2| + C$.

Correct Option: (B)

Question 23

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

$\int \frac{x}{x^2(x^2+1)} dx$. Let $x^2=t$.
$\frac{1}{2} \int \frac{dt}{t(t+1)} = \frac{1}{2}\log|\frac{t}{t+1}| = \frac{1}{2}\log|\frac{x^2}{x^2+1}|$.
$= \log|x| – \frac{1}{2}\log|x^2+1| + C$.

Correct Option: (A)