Accessing ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.10 can be a valuable tool for students seeking extra practice.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.10

Question 1.
(i) ∫ $$\frac{x-1}{(x-2)(x-3)}$$ dx
(ii) ∫ $$\frac{3 x+5}{x^2+3 x-18}$$ dx
Solution:
(i) Let $$\frac{x-1}{(x-2)(x-3)}$$ = $$\frac{\mathrm{A}}{x-2}$$ + $$\frac{\mathrm{B}}{x-3}$$ ……….(1)
Multiplying both sides of eqn. (1) by (x – 2) (x – 3) ; we have
x – 1 = A (x – 3) + B (x – 2) ……………(2)
putting x = 2, 3 successively in eqn. (2) ; we have
1 = – A
A = – 1
and 2 = B
∴ from (1); we have
$$\int \frac{x-1}{(x-2)(x-3)} d x=\int \frac{-1}{x-2} d x+\int \frac{2 d x}{x-3}$$
= – log |x – 2| + 2 log |x – 3| + C

(ii) Let $$\frac{3 x+5}{x^2+3 x-18}=\frac{3 x+5}{(x-3)(x+6)}$$
= $$\frac{\mathrm{A}}{x-3}+\frac{\mathrm{B}}{x+6}$$ …………..(1)
Multiplying both sides of eqn. (1) by (x – 3) (x + 6) ; we have
3x + 5 = A (x + 6) + B (x – 3) …………….(2)
putting x = 3, – 6 successively in eqn. (2) ;
14 = 9A
⇒ A = $$\frac{14}{9}$$
and – 13 = – 9B
⇒ B = $$\frac{13}{9}$$
∴ from (1) ;
$$\int \frac{3 x+5}{x^2+3 x-18} d x=\frac{14}{9} \int \frac{d x}{x-3}+\frac{13}{9} \int \frac{d x}{x+6}$$
= $$\frac{14}{9}$$ log |x – 3| + $$\frac{13}{9}$$ log |x – 6| + C

Question 1 (old).
(i) ∫ $$\frac{d x}{(x+1)(x+2)}$$ (NCERT)
(ii) ∫ $$\frac{x}{(x+1)(x+2)}$$ (NCERT)
Solution:
(i) Let $$\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$$ ………….(1)
Multiply both sides of eqn. (1) by (x + 1) (x + 2) ; we get
I = A (x + 2) + B (x + 1) ……………….(2)
putting x = – 1, – 2 successively in eqn. (2) ; we have
1 = A and 1 = – B
⇒ B = – 1
∴ from (1) ; we have
$$\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}$$
I = $$\frac{d x}{(x+1)(x+2)}$$
= ∫ $$\frac{d x}{x+1}$$ – ∫ $$\frac{d x}{x+2}$$
= log |x + 1| – log |x + 2| + C
= log $$\left|\frac{x+1}{x+2}\right|$$ + C

(ii) Let $$\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$$ …………..(1)
Multiplying eqn. (1) by (x + 1) (x + 2) ; we have
x = A (x + 2) + B (x + 1) ……………(2)
putting x = – 1 in eqn. (2) ;
– 1 = A
putting x = – 2 in eqn. (2) ; we have
– 2 = – B
⇒ B = 2
∴ eqn. (1) gives ;
∴ $$\int \frac{x}{(x+1)(x+2)} d x=-\int \frac{d x}{x+1}+2 \int \frac{d x}{x+2}$$
= – log |x + 1| + 2 log |x + 2| + C
= log $$\frac{(x+2)^2}{|x+1|}$$ + C

Question 2.
(i) ∫ $$\frac{2 x+7}{x^2-x-2}$$ dx (ISC 2020)
(ii) ∫ $$\frac{x^2+1}{x^2-5 x+6}$$ dx
Solution:
(i) Let $$\frac{2 x+7}{x^2-x-2}$$ = $$\frac{2 x+7}{(x-2)(x+1)}$$
= $$\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x+1}$$ …………….(1)
Multiply both sides of eqn. (1) by (x – 20 (x + 1) ; we have
2x + 7 = A (x + 1) + B (x – 2) ………….(2)
putting x = – 1 in eqn. (2) ; we have
11 = 3A
A = $$\frac{11}{3}$$
∴ from (1) ; we have
∫ $$\frac{2 x+7}{x^2-x-2}$$ dx = $$\int \frac{11 / 3}{x-2} d x+\int \frac{-5 / 3}{x+1} d x$$
= $$\frac{11}{3}$$ log |x – 2| – $$\frac{5}{3}$$ log |x – 1| + C

(ii) Let I = ∫ $$\frac{x^2+1}{x^2-5 x+6}$$ dx
= ∫ $$\left[1+\frac{5 x-5}{x^2-5 x+6}\right]$$ dx
= x + ∫ $$\frac{5(x-1) d x}{(x-2)(x-3)}$$
= x + 5I1
where I1 = ∫ $$\frac{(x-1) d x}{(x-2)(x-3)}$$
Let $$\frac{x-1}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}$$ ……………(1)
Multiplying both sides of eqn. (1) by (x – 2) (x – 3) ; we have
x – 1 = A (x – 3) + B (x – 2) …………..(2)
putting x = 2, 3 successively in eqn. (2)
1 = – A
⇒ A = – 1
and 2 = B
∴ from (1) ;
I1 = ∫ $$\frac{-1}{x-2}$$ dx + ∫ $$\frac{2}{x-3}$$ dx
= – log |x – 2| + 2 log |x – 3|
Thus I = x – 5 log |x – 2| + 10 log |x- 3| + C

Question 2 (old).
(i) ∫ $$\frac{x}{(x-1)(x-2)}$$ dx (NCERT)
(ii) ∫ $$\frac{x^2+1}{x^2-5 x+6}$$ dx (NCERT)
Solution:
(i) Let $$\frac{x}{(x-1)(x-2)}$$ = $$\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}$$ ……….(1)
Multiplying both sides eqn. (1) by (x – 1) (x – 2) ; we get
x = A (x – 2) + B (x – 1) ………….(2)
putting x = 1, 2 successively in eqn. (2) ; we have
1 = A (- 1)
⇒ A = – 1
and 2 = B
∴ from (1) ; we get
$$\frac{x}{(x-1)(x-2)}=-\frac{1}{x-1}+\frac{2}{x-2}$$
∴ ∫ $$\frac{x d x}{(x-1)(x-2)}$$ = – ∫ $$\frac{1}{x-1}$$ dx + 2 ∫ $$\frac{1}{x-2}$$ dx
= – log |x – 1| + 2 log |x – 2| + C

(ii) Let $$\frac{x^2+1}{(x-2)(x-3)}$$ = 1 + $$\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}$$ ……….(1)
Multiplying both sides eqn. (1) by (x – 2) (x – 3) ; we get
x2 + 1 = (x – 2) (x – 3) + A (x – 3) + B (x – 2) ………..(2)
putting x = 2 in eqn. (2) ;
– 5 = A
putting x = 3 in eqn. (2)
10 = B
∴ from (1) ; we have
$$\frac{x^2+1}{(x-2)(x-3)}$$ = 1 – $$\frac{5}{x-2}+\frac{10}{x-3}$$
∴ ∫ $$\frac{\left(x^2+1\right) d x}{(x-2)(x-3)}$$ = x – 5 log |x – 2| + 10 log |x – 3| + C

Question 3.
(i) ∫ $$\frac{x^3+x+1}{x^2-1}$$ dx (NCERT)
(ii) ∫ $$\frac{x^3+1}{x^3-x}$$ dx
Solution:
(i) Let $$\frac{x^3+x+1}{x^2-1}$$ = x + $$\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+1}$$ ………….(1)
Multiply eqn. (1) by (x2 – 1) ; we get
x3 + x + 1 = x (x2 – 1) + A (x + 1) B (x – 1) …………..(2)
putting x = 1 in eqn. (2) ;
3 = 2A
⇒ A = $$\frac{3}{2}$$
putting x = – 1 in eqn. (2) ;
– 1 = – 2B
⇒ B = $$\frac{1}{2}$$
∴ from (1) ; we have
$$\frac{x^3+x+1}{x^2-1}=x+\frac{\frac{3}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$$
∴ ∫ $$\frac{x^3+x+1}{x^2+1}$$ dx = $$\frac{x^2}{2}$$ + $$\frac{3}{2}$$ log |x – 1| + $$\frac{1}{2}$$ log |x + 1| + C

(ii) Let I = ∫ $$\frac{x^3+1}{x^3-x}$$ dx
= ∫ $$\frac{x^3-x+x+1}{x^3-x}$$ dx
= ∫ $$\left[1+\frac{x+1}{x\left(x^2-1\right)}\right]$$ dx
= ∫ $$\left[1+\frac{x+1}{x(x-1)(x+1)}\right]$$ dx
= ∫ $$\left[1+\frac{1}{x(x-1)}\right]$$ dx
= ∫ $$\left[1+\frac{-1}{x}+\frac{1}{x-1}\right]$$ dx
= x – log |x| + log |x – 1| + C

Question 3 (old).
(ii) ∫ $$\frac{1-x^2}{x(1-2 x)}$$ dx (NCERT)
Solution:
Let $$\frac{1-x^2}{x(1-2 x)}$$ = $$\frac{1}{2}+\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-2 x}$$ ………….(1)
Multiply eqn. (1) by x (1 – 2x) ;
1 – x2 = $$\frac{1}{2}$$ x (1 – 2x) + A (1 – 2x) + B
putting x = 0 in eqn. (2) ;
1 = A
putting x = $$\frac{1}{2}$$ in eqn. (2) ;
$$\frac{3}{4}=\frac{B}{2}$$
B = $$\frac{3}{2}$$
∴ eqn. (1) gives ;
$$\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{x}+\frac{\frac{3}{2}}{1-2 x}$$
∴ ∫ $$\frac{\left(1-x^2\right)}{x(1-2 x)}$$ dx = $$\frac{1}{2}$$ x + log |x| + $$\frac{3}{2} \log \frac{|1-2 x|}{-2}$$ + C
= $$\frac{x}{2}$$ + log |x| – $$\frac{3}{4}$$ log |1 – 2x| + C

Question 4.
(i) ∫ $$\frac{x}{(x-1)(x-2)(x-3)}$$ dx (NCERT)
(ii) ∫ $$\frac{3 x-1}{(x-1)(x-2)(x-3)}$$ dx (NCERT)
Solution:
(i) Let $$\frac{x}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}$$ ………(1)
Multiplying eqn. (1) by (x – 1) (x – 2) (x – 3) ; we get
x = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) …………(2)
putting x = 1 in eqn. (2) ;
1 = 2 A
⇒ A = $$\frac{1}{2}$$
putting x = 1 in eqn. (2) ;
2 = – B
⇒ B = – 2
putting x = 3 in eqn. (2) ;
3 = 2 C
⇒ C = $$\frac{3}{2}$$
∴ eqn. (1) becomes,
$$\frac{x}{(x-1)(x-2)(x-3)}=\frac{\frac{1}{2}}{x-1}-\frac{2}{x-2}+\frac{\frac{3}{2}}{x-3}$$
∴ $$\int \frac{x d x}{(x-1)(x-2)(x-3)}=\frac{1}{2} \int \frac{d x}{x-1}-2 \int \frac{d x}{x-2}+\frac{3}{2} \int \frac{d x}{x-3}$$
= $$\frac{1}{2}$$ log |x – | – 2 log |x – 2| + $$\frac{3}{2}$$ log |x – 3| + C

(ii) Let $$\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}$$ ……….(1)
Multiplying eqn. (1) by (x – 1) (x – 2) (x – 3) ; we have
3x – 1 = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) …………..(2)
putting x = 1 in eqn. (2) ;
2 = 2A
⇒ A = 1 ;
putting x = 2 in eqn. (2) ;
5 = – B
⇒ B = – 5
putting x = 3 in eqn. (2) ;
8 = 2C
⇒ C = 4
∴ eqn. (1) gives ;
$$\int \frac{(3 x-1) d x}{(x-1)(x-2)(x-3)}=\int \frac{d x}{x-1}-\int \frac{5 d x}{x-2}+4 \int \frac{d x}{x-3}$$
= log |x – 1| – 5 log |x – 2| + 4 log |x – 3| + C

Question 5.
(i) ∫ $$\frac{5 x}{(x+1)\left(x^2-4\right)}$$ dx
(ii) ∫ $$\frac{x}{(x-1)^2(x+2)}$$ dx (NCERT)
Solution:
(i) Let $$\frac{5 x}{(x+1)\left(x^2-4\right)}$$ = $$\frac{5 x}{(x+1)(x-2)(x+2)}$$
i.e. $$\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x+2}$$ ……………..(1)
Multiplying eqn. (1) by (x + 1) (x – 2) (x + 2) ; we get
5x = A (x2 – 4) + B (x + 1) (x + 2) + C (x + 1) (x – 2) ………..(2)
putting x = – 1 in eqn. (2) ;
– 5 = – 3A
⇒ A = $$\frac{5}{3}$$
putting x = 2 in eqn. (2) ;
10 = 12 B
⇒ B = $$\frac{5}{6}$$
putting x = – 2 in eqn. (2) ;
– 10 = 4C
⇒ C = – $$\frac{5}{2}$$
∴ eqn. (1) gives ;
$$\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{\frac{5}{3}}{x+1}+\frac{\frac{5}{6}}{x-2}-\frac{\frac{5}{2}}{x+2}$$
∴ ∫ $$\frac{5 x d x}{(x+1)\left(x^2-4\right)}$$ = $$\frac{5}{3} \int \frac{d x}{x+1}+\frac{5}{6} \int \frac{d x}{x-2}-\frac{5}{2} \int \frac{d x}{x+2}$$
= $$\frac{5}{3}$$ log |x + 1| + $$\frac{5}{6}$$ log |x – 2| – $$\frac{5}{2}$$ log |x + 2| + C

(ii) Let $$\frac{x}{(x-1)^2(x+2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{x+2}$$ ………….(1)
Then x = A (x – 1) (x + 2) + B (x + 2) + C (x – 1)2 …………..(2)
putting x = 1, – 2, and 0 successively in eqn. (2) ; we get
1 = 3B
⇒ B = 1/3
– 2 = 9C
⇒ C = – 2/9
and 0 = – 2A + 2B + 2C
⇒ 2A = $$\frac{2}{3}-\frac{2}{9}$$
A = 2/9
From (1) ; we have

Question 6.
(i) ∫ $$\frac{3 x-1}{(x-2)^2}$$ dx
(ii) ∫ $$\frac{3 x+5}{x^3-x^2-x+1}$$ dx (NCERT)
Solution:
(i) Let $$\frac{3 x-1}{(x-2)^2}$$ = $$\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{(x-2)^2}$$ ………..(1)
Multiply both sides of eqn. (1) by (x – 2)2 ; we have
3 = A ;
– 1 = – 2A + B
⇒ B = 2
∴ from (1) ;
$$\int \frac{3 x-1}{(x-2)^2} d x=\int \frac{3}{x-2} d x+\int \frac{5}{(x-2)^2} d x$$
= 3 log |x – 2| + 5 $$\frac{(x-2)^{-2+1}}{(-2+1)}$$ + C
= 3 log |x – 2| – $$\frac{5}{x-2}$$ + C

(ii) $$\frac{3 x+5}{x^3-x^2-x+1}=\frac{3 x+5}{(x-1)\left(x^2-1\right)}=\frac{3 x+5}{(x-1)^2(x+1)}$$
Let $$\frac{3 x+5}{(x-1)^2(x+1)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-1}+\frac{\mathrm{C}}{(x-1)^2}$$ ……………..(1)
Multiplying eqn. (1) by (x – 1)2 (x + 1) ; we get
3x + 5 = A (x – 1)2 + B (x2 – 1) + C (x + 1) ……………(2)
putting x = 1 in eqn. (2) ;
8 = 2C
⇒ C = 4
putting x = – 1 in eqn. (2) ;
2 = 4A
⇒ A = $$\frac{1}{2}$$
Coeff. of x2 ;
0 = A + B
⇒ B = – $$\frac{1}{2}$$
∴ eqn. (1) gives ;
$$\int \frac{(3 x+5) d x}{x^3-x^2-x+1}=\frac{1}{2} \int \frac{d x}{x+1}-\frac{1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^2}$$
= $$\frac{1}{2}$$ log |x + 1| – $$\frac{1}{2}$$ log |x – 1| – $$\frac{4}{x-1}$$ + C
= $$\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}$$ + C, x ≠ 1.

Question 7.
(i) ∫ $$\frac{2}{(1-x)\left(1+x^2\right)}$$ dx
(ii) ∫ $$\frac{5 x}{(x+1)\left(x^2+9\right)}$$ dx (NCERT)
(iii) ∫ $$\frac{4}{(x-2)\left(x^2+4\right)}$$ dx
Solution:
(i) Let $$\frac{2}{(1-x)\left(1+x^2\right)}=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}$$ …………(1)
Multiplying eqn. (1) by (1 – x) (1 + x2) ; we get
2 = A (1 + x2) + (Bx + C) (1 – x)
put x = 1 in eqn. (2) ;
2 = 2A
⇒ A = 1
Coeff. of x2 ;
0 = A – B
⇒ B = 1
Coeff. of x ;
0 = B – C
⇒ C = 1
∴ From (1) ; we get
$$\int \frac{2 d x}{(1-x)\left(1+x^2\right)}=\int \frac{d x}{1-x}+\int \frac{(x+1)}{x^2+1} d x$$
= $$\frac{\log |1-x|}{-1}+\frac{1}{2} \int \frac{2 x d x}{x^2+1}+\int \frac{d x}{x^2+1}$$
= – log |1 – x| + $$\frac{1}{2}$$ log |x2 + 1| + tan-1 x + C

(ii) Let $$\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+9}$$ …………(1)
Multiplying eqn. (1) by (x + 1) (x2 + 9) ; we get
5x = A (x2 + 9) + (Bx + C) (x + 1)
putting x = – 1 in eqn. (2) ; we get
– 5 = 10A
⇒ A = – $$\frac{1}{2}$$
Coeff. of x2 ;
0 = A + B
⇒ B = $$\frac{1}{2}$$
Coeff. of x ;
5 = B + C
⇒ C = $$\frac{9}{2}$$
∴ From eqn. (1) ; we have

(iii) Let I = ∫ $$\frac{4}{(x-2)\left(x^2+4\right)}$$ dx
Let $$\frac{4}{(x-2)\left(x^2+4\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2+4}$$ …………(1)
Multiply both sides of eqn. (1) by (x – 2) (x2 + 4) ; we have
4 = A (x2 + 4) + (Bx + C) (x – 2) …………….(2)
put x = 2 in eqn. (2) ; we have
4 = 8A
⇒ A = $$\frac{1}{2}$$
Coeff. of x2 ;
0 = A + B
⇒ B = $$\frac{1}{2}$$
Coeff. of x ;
0 = – 2B + C
⇒ C = 2B
= 2 (- $$\frac{1}{2}$$) = – 1
∴ From eqn. (1) ; we have
I = $$\int \frac{\frac{1}{2}}{x-2} d x+\int \frac{-\frac{x}{2}-1}{x^2+4} d x$$
= $$\frac{1}{2}$$ log |x – 2| – $$\frac{1}{2} \int \frac{x d x}{x^2+4}-\int \frac{d x}{x^2+4}$$ + C
= $$\frac{1}{2} \log |x-2|-\frac{1}{4} \log \left|x^2+4\right|-\frac{1}{2} \tan ^{-1} \frac{x}{2}$$ + C
[∵ ∫ $$\frac{f^{\prime}(x)}{f(x)}$$ dx = log |f(x)| + C
and ∫ $$\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]

Question 7 (old).
(i) ∫ $$\frac{x}{(x-1)\left(x^2+1\right)}$$ dx (NCERT)
(ii) ∫ $$\frac{x^2+x+1}{(x+2)\left(x^2+1\right)}$$ dx (NCERT)
Solution:
(i) Let $$\frac{x}{(x-1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$$ …………(1)
Multiply both sides of eqn. (1) by (x – 1) (x2 + 1) ; we have
x = A (x2 + 1) + (Bx + C) (x – 1)
putting x = 1 in eqn. (2) ; we have
1 = 2A
⇒ A = $$\frac{1}{2}$$
Coefficients of x2 ;
0 = A + B
⇒ B = $$-\frac{1}{2}$$
Coefficients of x ;
1 = – B + C
⇒ C = 1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$
∴ From eqn. (1) ; we have

(ii) Let $$\frac{x^2+x+1}{(x+2)\left(x^2+1\right)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$$ ………..(1)
Multiplying both sides of eqn. (1) by (x2 + 1) (x + 2) ; we get
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2)
putting x = – 2 in eqn. (2) ; we have
4 – 2 + 1 = A (4 + 1)
⇒ A = $$\frac{3}{5}$$
Coefficients of x2 ;
1 = A + B
⇒ B = 1 – $$\frac{3}{5}$$ = $$\frac{2}{5}$$
Coefficients of x ;
1 = 2B + C
⇒ C = 1 – $$\frac{4}{5}$$ = $$\frac{1}{5}$$
∴ From eqn. (1) ; we have

Question 8.
(i) ∫ $$\frac{x^2+x}{x^3-x^2+x-1}$$ dx
(ii) ∫ $$\frac{x^3}{(x-1)\left(x^2+1\right)}$$ dx
Solution:
(i) Let $$\frac{x^2+x}{x^3-x^2+x-1}=\frac{x^2+x}{\left(x^2+1\right)(x-1)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$$ ……………….(1)
Multiply both sides of eqn. (1) by (x2 + 1) (x – 1) ; we have
x2 + x = A (x2 + 1) + (Bx + C) (x – 1) …………….(2)
putting x = 1 in eqn. (2) ; we have
1 + 1 = A (1 + 1)
⇒ A = 1
Coefficients of x2 ;
1 = A + B
⇒ B = 0
Coefficients of x ;
1 = – B + C
⇒ C = 1
∴ from (1) ; we get
$$\frac{x^2+x}{x^3-x^2+x-1}=\frac{1}{x-1}+\frac{1}{x^2+1}$$
∴ $$\int \frac{\left(x^2+x\right) d x}{x^3-x^2+x-1}=\int \frac{1}{x-1} d x+\int \frac{d x}{x^2+1}$$
= log |x – 1| + tan-1 x + C
[∵ ∫ $$\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]

(ii) Let I = ∫ $$\frac{x^3}{(x-1)\left(x^2+1\right)}$$
Here the integrand is not a proper function.
So on dividing x3 by (x – 1) (x2 + 1) ; we get
quotient = 1
and remainder = x2 – x + 1
∴ $$\frac{x^3}{(x-1)\left(x^2+1\right)}=1+\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}$$ …………..(1)
Let $$\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}$$ …………….(2)
Multiplying both sides of eqn. (1) by (x – 1) (x2 + 1) ; we get
x2 – x + 1 = A (x2 + 1) + (Bx + C) (x – 1) ……………(3)
putting x = 1 in eqn. (3) ; we have
1 – 1 + 1 = A (2)
⇒ A = $$\frac{1}{2}$$
Coefficients of x2 ;
1 = A + B
⇒ B = $$\frac{1}{2}$$
Coefficients of x ;
– 1 = – B + C
⇒ C = – 1 + $$\frac{1}{2}$$ = – $$\frac{1}{2}$$

Question 9.
(i) ∫ $$\frac{d x}{1-x^3}$$
(ii) ∫ $$\frac{d x}{(x+1)^2\left(x^2+1\right)}$$
Solution:
(i) $$\frac{1}{1-x^3}=\frac{1}{(1-x)\left(1+x+x^2\right)}$$
Let $$\frac{1}{(1-x)\left(1+x+x^2\right)}=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+x+1}$$ ………….(1)
Multiplying eqn. (1) by (1 – x) (x2 + x + 1) ; we get
1 = A (x2 + x + 1) + (Bx + C) (1 – x) ………….(2)
putting x = 1 in eqn. (2);
1 = 3A
⇒ A = $$\frac{1}{3}$$
Coefficients of x2 ;
0 = A – B
⇒ B = $$\frac{1}{3}$$
Coefficients of x ;
0 = A + B – C
⇒ C = $$\frac{2}{3}$$
From (1) ; we have

Question 10.
(i) ∫ $$\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}$$ dx
(ii) ∫ $$\frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)}$$ dx
(iii) ∫ $$\frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}$$ dx
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ $$\frac{d t}{(t+1)(t+3)}$$
Let $$\frac{1}{(t+1)(t+3)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+3}$$ …………….(1)
⇒ 1 = A (t + 3) + B (t + 1) ………….(2)
putting t = – 1, – 3 successively in eqn. (2) ; we get
1 = 2A
⇒ A = 1/2
and 1 = – 2B
⇒ B = – 1/2
∴ from (1) ;
$$\frac{1}{(t+1)(t+3)}=\frac{1 / 2}{t+1}+\frac{-1 / 2}{t+3}$$
∴ I = $$\int \frac{\frac{1}{2}}{t+1} d t-\frac{1}{2} \int \frac{d t}{t+3}$$
= $$\frac{1}{2}$$ log |t + 1| – $$\frac{1}{2}$$ log |t + 3| + c
= $$\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|$$ + c
= $$\frac{1}{2} \log \left|\frac{x^2+1}{x^2+3}\right|$$ + c

(ii) Let I = ∫ $$\frac{x^2 d x}{\left(1+x^3\right)\left(2+x^3\right)}$$ ;
put x3 = y
⇒ 3x2 dx = dy
= $$\frac{1}{3} \int \frac{d y}{(y+1)(y+2)}$$ …………..(1)
Let $$\frac{1}{(y+1)(y+2)}=\frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+2}$$ ………….(2)
Multiplying eqn. (2) by (y + 1) (y + 2) ; we get
1 = A (y + 2) + B (y + 1) ……………..(3)
putting y = – 1 in eqn. (3) ;
1 = A
putting y = – 2in eqn. (3) ;
1 = – B
⇒ B = – 1
∴ From (2) ; we have
$$\frac{1}{(y+1)(y+2)}=\frac{1}{y+1}-\frac{1}{y+2}$$
∴ $$\int \frac{d y}{(y+1)(y+2)}=\int \frac{d y}{y+1}-\int \frac{d y}{y+2}$$
∴ From (1) ; we have
1 = $$\frac{1}{3}$$ [log |y + 1| – kog |y + 2|] + C
= $$\frac{1}{3} \log \left|\frac{x^3+1}{x^3+2}\right|$$ + C

(iii) Let I = ∫ $$\frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)^2}$$ dx
put x2 = t
⇒ 2x dx = dt
= ∫ $$\frac{d t}{(t+1)(t+2)^2}$$
Let $$\frac{1}{(t+1)(t+2)^2}$$ = $$\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2}+\frac{\mathrm{C}}{(t+2)^2}$$ …………….(1)
Multiply both sides of eqn. (1) by (t + 1) (t + 2)2 ; we have
1 = A (t + 2)2 + B (t + 1) (t + 2) + C (t + 1) ………….(2)
putting t = – 1 in eqn. (2) ; we have
1 = A (1)2
⇒ A = 1
putting t = – 2 in eqn. (2) ; we have
1 = C (- 2 + 1)
⇒ C = – 1
Coeff. of t2 ;
0 = A + B
⇒ B = – A = – 1
∴ from (1) ; we have
∴ I = ∫ $$\frac{1}{(t+1)(t+2)^2}$$ dt
= $$\int \frac{d t}{t+1}-\int \frac{d t}{t+2}-\int \frac{d t}{(t+2)^2}$$
= log |t + 1| – log |t + 2| – $$\frac{(t+2)^{-2+1}}{-2+1}$$ + C
= log (x2 + 1) – log (x2 + 2) + $$\frac{1}{x^2+2}$$ + C

Question 11.
(i) ∫ $$\frac{d x}{x\left(x^4-1\right)}$$ (NCERT)
(ii) ∫ $$\frac{d x}{x\left(x^2+1\right)}$$ (NCERT)
Solution:
(i) Let I = ∫ $$\frac{d x}{x\left(x^4-1\right)}$$
put x4 = t
⇒ 4x3 dx = dt
⇒ dx = $$\frac{d t}{4 x^3}$$
∴ I = ∫ $$\frac{d t}{4 x^4\left(x^4-1\right)}$$
= $$\frac{1}{4} \int \frac{d t}{t(t-1)}$$ …………….(1)
Let $$\frac{1}{t(t-1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t-1}$$ ……………(2)
⇒ 1 = A (t – 1) + Bt …………..(3)
putting t = 0, 1 successively in eqn. (3) ; we have
1 = – A
⇒ A = – 1
and 1 = B
⇒ B = 1
∴ From (2) ;
$$\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$$
∴ from eqn. (1) ; we have
1 = $$\frac{-1}{4}$$ log |t| + $$\frac{1}{4}$$ log |t – 1| + c
= $$\frac{1}{4} \log \left|\frac{t-1}{t}\right|$$ + c
= $$\frac{1}{4} \log \left|\frac{x^4-1}{x^4}\right|$$ + c

(ii) Let I = ∫ $$\frac{d x}{x\left(x^2+1\right)}$$
= ∫ 
put x2 dx = t
⇒ 2x dx = dt
∴ I = ∫ $$\frac{d t}{2 t(t+1)}$$
Let $$\frac{1}{t(t+1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+1}$$ …………..(1)
Multiplying both sides of eqn. (1) by t (t + 1) ; we have
1 = A (t + 1) + Bt ……………..(2)
putting t = 0, – 1 successively in eqn. (2) ; we have
1 = A
and 1 = – B
B = – 1
∴ from (1) ;

Question 12.
(i) ∫ $$\frac{d x}{x\left(x^n+1\right)}$$ (NCERT)
(ii) ∫ $$\frac{d x}{x\left(x^3+1\right)}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{x\left(x^n+1\right)}$$
putting xn = t
n xn-1 dx = dt
⇒ dx = $$\frac{d t}{n x^{n-1}}$$
∴ I = ∫ $$\frac{d t}{n x^n\left(x^n+1\right)}$$
= $$\frac{1}{n} \int \frac{d t}{t(t+1)}$$
Let $$\frac{1}{t(t+1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+1}$$ ………….(1)
⇒ 1 = A (t + 1) + Bt ………..(2)
putting t = 0, – 1 successively in eqn. (2) ; we have
∴ From (1) ;
$$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$$
Thus I = $$\frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right]$$
= $$\frac{1}{n}$$ [log |t| – log |t + 1| + c]
= $$\frac{1}{n} \log \left|\frac{t}{t+1}\right|$$ + c
= $$\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|$$ + c

(ii) Let I = ∫ $$\frac{d x}{x\left(x^3+8\right)}$$
putting x3 = t
⇒ 3x2 dx = dt
⇒ dx = $$\frac{d t}{3 x^2}$$
∴ I = ∫ $$\frac{d t}{3 x^3\left(x^3+8\right)}$$
= $$\frac{1}{3} \int \frac{d t}{t(t+8)}$$
Let $$\frac{1}{t(t+8)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+8}$$ …………….(1)
⇒ 1 = A (t + 8) + Bt …………..(2)
putting t = 0, – 8 successively in eqn. (2) ; we have
1 = 8A
⇒ A = 1/8
and 1 = – 8B
⇒ B = – 1/8
∴ From (1) ;

Question 13.
(i) ∫ $$\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}$$ dx (NCERT)
(ii) ∫ $$\frac{d x}{e^{2 x}+1}$$
Solution:
(i) put ex = t
⇒ ex dx = dt
∴ I = ∫ $$\frac{d t}{(1+t)(2+t)}$$
Let $$\frac{1}{(1+t)(2+t)}=\frac{\mathrm{A}}{1+t}+\frac{\mathrm{B}}{2+t}$$ ………….(1)
Multiplying eqn. (1) by (1 + t) (2 + t) ; we have
1 = A (2 + t) + B (1 + t) ………..(2)
put t = – 1 in eqn. (2) ;
⇒ 1 = A
put t = – 2 in eqn. (2) ;
1 = – B
⇒ B = – 1
∴ I = $$\int \frac{d t}{1+t}-\int \frac{d t}{2+t}$$
= log |1 + t| – log |2 + t| + C
= log $$\left|\frac{1+t}{2+t}\right|$$ + C
= log $$\left|\frac{1+e^x}{2+e^x}\right|$$ + C

(ii) I = ∫ $$\frac{d x}{e^{2 x}+1}$$
put ex = t
⇒ ex dx = dt
⇒ dx = $$\frac{d t}{t}$$
∴ I = ∫ $$\frac{d t}{t\left(t^2+1\right)}$$
Let $$\frac{1}{t\left(t^2+1\right)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B} t+\mathrm{C}}{t^2+1}$$ ………….(1)
Multiplying both sides of eqn. (1) by t (t2 + 1) ; we have
1 = A (t2 + 1) + (Bt + C) t …………….(2)
putting t = 0 in eqn. (2) ; we have
1 = A
Coeff. of t2 ;
0 = A + B
⇒ B = – 1
Coeff. of t ;
0 = C
∴ from (1) ;
$$\frac{1}{t\left(t^2+1\right)}=\frac{1}{t}-\frac{t}{t^2+1}$$
Thus, I = ∫ $$\left[\frac{1}{t}-\frac{t}{t^2+1}\right]$$ dt

Question 14.
(i) ∫ $$\frac{\cos x}{(1-\sin x)(3-\sin x)}$$ dx
(ii) ∫ $$\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}$$ dx (ISC 2017)
Solution:
(i) Let I = ∫ $$\frac{\cos x}{(1-\sin x)(3-\sin x)}$$ dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ $$\frac{d t}{(1-t)(3-t)}$$
Let $$\frac{1}{(1-t)(3-t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{3-t}$$ ……………(1)
Multiplying both sides of eqn. (1) by (1 – t) (3 – t) ; we have
1 = A (3 – t) + B (1 – t) ………..(2)
putting t = 1, 3 successively in eqn. (2) ; we have
1 = 2A
⇒ A = $$\frac{1}{2}$$
and 1 = – 2B
⇒ B = – $$\frac{1}{2}$$

(ii) Let I = ∫ $$\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}$$ dx
= ∫ $$\frac{2 \sin x \cos x d x}{(1+\sin x)(2+\sin x)}$$
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{2 t d t}{(1+t)(2+t)}$$
Let $$\frac{2 t}{(t+1)(t+2)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2}$$ ………….(1)
Multiplying both sides of eqn. (1) by (t + 1) (t + 2) ; we have
2t = A (t + 2) + B (t + 1)
putting t = – 1, – 2 successively in eqn. (2) ; we get
– 2 = A
and – 4 = – B
⇒ B = 4
∴ from (1) ;
I = ∫ $$\left[\frac{-2}{t+1}+\frac{4}{t+2}\right]$$ dt
= – 2 log |1 + t| + 4 log |2 + t| + C
= – 2 log |1 + sin x| + 4 log |2 + sin x| + C

Question 15.
(i) ∫ $$\frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)}$$ dx
(ii) ∫ $$\frac{\sin 2 x}{\left(\sin ^2 x+1\right)\left(\sin ^2 x+3\right)}$$ dx
Solution:
(i) Let I = ∫ $$\frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)}$$ dx
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{2 d t}{(1-t)\left(1+t^2\right)}$$
Let $$\frac{2}{(1-t)\left(1+t^2\right)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B} t+\mathrm{C}}{1+t^2}$$ …………..(1)
Multiplying both sides of eqn. (1) by (1 – t) (1 + t2) ; we get
2 = A (1 + t2) + (Bt + C) (1 – t) …………(2)
2 = 2A
⇒ A = 1
putting t = 1 in eqn. (2) ; we have
Coeff. of t2 ;
0 = A – B
⇒ B = 1
Coeff. of t ;
0 = B – C
⇒ C = B = 1
∴ from eqn. (1) ; we have

(ii) Let I = ∫ $$\frac{\sin 2 x}{\left(\sin ^2 x+1\right)\left(\sin ^2 x+3\right)}$$ dx
put sin2 x = t
⇒ 2 sin x cos x dx = dt
⇒ sin 2x dx = dt
∴ I = ∫ $$\frac{d t}{(t+1)(t+3)}$$
Let $$\frac{1}{(t+1)(t+3)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+3}$$ ……………..(1)
Multiplying both sides of eqn. (1); we have
1 =A (t + 3) +B (t + 1)
putting t = – 3 in eqn. (2); we have
1 = – 2B
⇒ B = – $$\frac{1}{2}$$
putting t = – 1 in eqn. (2); we have
1 = 2A
⇒ A = $$\frac{1}{2}$$
∴ from (1); we have
I = ∫ $$\left[\frac{\frac{1}{2}}{t+1}-\frac{\frac{1}{2}}{t+3}\right]$$ dt
= $$\frac{1}{2}$$ log |t + 1| – $$\frac{1}{2}$$ log |t + 3| + C
I = $$\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|$$ + C

Question 15 (old).
(ii) ∫ $$\frac{\sin 2 x}{(1-\cos 2 x)(2-\cos 2 x)}$$ dx
(iii) ∫ $$\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}$$ dx
Solution:
(ii) Let I = ∫ $$\frac{\sin 2 x}{(1-\cos 2 x)(2-\cos 2 x)}$$ dx
put cos 2x = t
⇒ – 2 sin 2x dx = dt
∴ I = – $$\frac{1}{2} \int \frac{d t}{(1-t)(2-t)}$$
Let $$\frac{1}{(1-t)(2-t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{2-t}$$ ………………(1)
Multiplying eqn. (1) by (1 – t) (2 – t) ; we have
1 = A (2 – t) + B (1 – t) …………..(2)
put t = 1 in eqn. (2) ;
1 = A
put t = 2 in eqn. (2) ;
1 = – B
⇒ B = – 1
∴ From (1) ; we get
$$\frac{1}{(1-t)(2-t)}=\frac{1}{1-t}-\frac{1}{2-t}$$
∴ I = – $$\frac{1}{2}\left[\int \frac{d t}{1-t}-\int \frac{d t}{2-t}\right]$$
= – $$\frac{1}{2}\left[\frac{\log |1-t|}{-1}-\frac{\log |2-t|}{-1}\right]$$ + C
= $$\frac{1}{2} \log \left|\frac{1-\cos 2 x}{2-\cos 2 x}\right|$$ + C

(iii) Let I = ∫ $$\frac{\sin 2 x d x}{(1+\sin x)(2+\sin x)}$$
= ∫ $$\frac{2 \sin x \cos x d x}{(1+\sin x)(2+\sin x)}$$
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{2 t d t}{(1+t)(2+t)}$$
Let $$\frac{2 t}{(t+1)(2+t)}=\frac{\mathrm{A}}{1+t}+\frac{\mathrm{B}}{2+t}$$ ……………(1)
MuItpIying both sides of eqn. (1) by (1 + t) (2 + t) ; we get
2t = A (2 + t) + B (1 + t) ……………(2)
putting t = – 1 in eqn. (2) ; we have
⇒ – 2 = A
putting t = 2 in eqn. (2) ; we have
– 4 = – B
⇒ B = 4
∴ from (1) ;
I = ∫ $$\frac{2 t d t}{(1+t)(2+t)}$$
= $$\int \frac{-2}{1+t} d t+\int \frac{4 d t}{2+t}$$
= – 2 log |1 + t| + 4 log |2 + t| + C
= – 2 log |1 + sin x| + 4 log |2 + sin x| + C

Question 16.
(i) ∫ $$\frac{\sec ^2 x}{(2+\tan x)(3+\tan x)}$$ dx
(ii) ∫ $$\frac{d x}{\sin x(3+2 \cos x)}$$
(i) Let I = ∫ $$\frac{\sec ^2 x}{(2+\tan x)(3+\tan x)}$$ dx
put tan x = t
⇒ sec2 x dx = dt
∴ I = ∫ $$\frac{d t}{(2+t)(3+t)}$$
Let $$\frac{1}{(2+t)(3+t)}=\frac{\mathrm{A}}{2+t}+\frac{\mathrm{B}}{3+t}$$ ………….(1)
Multiply both sides of eqn. (1) by (2 + t) (3 + t) ; we have
1 = A (3 + t) + B (2 + t)
putting t = – 2, – 3 successively in eqn. (2) ; we have
1 = A
and – 1 = B
∴ from (1) ;
$$\frac{1}{(2+t)(3+t)}=\frac{1}{2+t}-\frac{1}{3+t}$$
Thus, I = ∫ $$\left[\frac{1}{2+t}-\frac{1}{3+t}\right]$$ dt
= log |2 + t| – log |3 + t|
= log $$\left|\frac{2+t}{3+t}\right|$$ + C
= log $$\left|\frac{2+\tan x}{3+\tan x}\right|$$ + C

(ii) Let I = ∫ $$\frac{d x}{\sin x(3+2 \cos x)}$$
= ∫ $$\frac{\sin x d x}{\left(1-\cos ^2 x\right)(3+2 \cos x)}$$
put cos x = t
– sin x dx = dt
∴ I = – ∫ $$\frac{d t}{(1-t)(1+t)(3+2 t)}$$ ……………(1)
Let $$\frac{1}{(1-t)(1+t)(3+2 t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}+\frac{\mathrm{C}}{3+2 t}$$ …………..(2)
Multiply both sides of eqn. (2) by (1 – t) (1 + t) (3 + 2t) ; we have
1 = A (1 + t) (3 + 2t) + B (1 – t) (3 + 2t) + C (1 – t) (1 + t)
putting t = 1, – 1, – $$\frac{3}{2}$$ successively in eqn. (3) ; we get
1 = 10A
⇒ A = $$\frac{1}{10}$$ ;
1 = 2B
⇒ B = $$\frac{1}{2}$$
and 1 = C $$\left(\frac{5}{2}\right)\left(\frac{-1}{2}\right)$$
⇒ C = – $$\frac{4}{5}$$
∴ from (1) ; we get
I = – $$\left[\int \frac{1 / 10}{1-t} d t+\int \frac{1 / 2 d t}{1+t}+\int \frac{-4 / 5 d t}{3+2 t}\right]$$
= – $$\left[\frac{-1}{10} \log |1-t|+\frac{1}{2} \log |1+t|-\frac{4}{5} \frac{\log |3+2 t|}{2}\right]$$ + C
= $$\frac{1}{10}$$ log |1 – cos x| – $$\frac{1}{2}$$ log |1 + cos x| + $$\frac{2}{5}$$ log |3 + 2 cos x| + C.

Question 17.
(i) ∫ $$\frac{d x}{\sin x-\sin 2 x}$$
(ii) ∫ $$\frac{d x}{x(6 \log x)^2+(7 \log x+2)}$$ (ISC 2012)
Solution:
(i) Let I = ∫ $$\frac{d x}{\sin x-\sin 2 x}$$
= ∫ $$\frac{d x}{\sin x(1-2 \cos x)}$$
put cos x = t
⇒ – sin x dx = dt
= ∫ $$\frac{\sin x d x}{\left(1-\cos ^2 x\right)(1-2 \cos x)}$$
= ∫ $$\frac{-d t}{\left(1-t^2\right)(1-2 t)}$$
Let $$\frac{-1}{\left(1-t^2\right)(1-2 t)}=\frac{-1}{(1-t)(1+t)(1-2 t)}$$
= $$\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}+\frac{\mathrm{C}}{1-2 t}$$ …………..(1)
Multiplying both sides of eqn. (1) by (1 – t) (1 + t) (1 – 2t) ; we have
– 1 = A (1 + t) (1 – 2t) + B (1 – t) (1 – 2t) + C (1 – t) (1 + t) …………..(2)
putting t = 1, – 1 and $$\frac{1}{2}$$ successively in eqn. (2) ; we have
– 1 = A (2) (- 1)
⇒ A = $$\frac{1}{2}$$
– 1 = B (2) (3)
⇒ B = – $$\frac{1}{6}$$
and – 1 = C $$\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)$$
⇒ C = – $$\frac{4}{3}$$
∴ from eqn. (1) ; we have

(ii) Let I = ∫ $$\frac{d x}{x\left[6(\log x)^2+7 \log x+2\right]}$$
put log x = t
⇒ $$\frac{1}{x}$$ dx = dt
= ∫ $$\frac{d t}{6 t^2+7 t+2}$$
= ∫ $$\frac{d t}{(2 t+1)(3 t+2)}$$
Let $$\frac{1}{(2 t+1)(3 t+2)}=\frac{\mathrm{A}}{2 t+1}+\frac{\mathrm{B}}{3 t+2}$$ …………(1)
Multiplying both sides of eqn. (1) by (2t + 1) (3t + 2) ; we have
⇒ 1 = A (3t + 2) + B (2t + 1) ……………..(2)
putting t = – $$\frac{2}{3}$$ in eqn. (2) ; we have
1 = B (- $$\frac{1}{3}$$)
⇒ B = – 3
putting t = – $$\frac{1}{2}$$ in eqn. (2) ; we have
1 = A (latex]\frac{1}{2}[/latex])
⇒ A = 2
∴ from (1) ;
I = ∫ $$\left[\frac{2}{2 t+1}-\frac{3}{3 t+2}\right]$$ dt
= $$\frac{2 \log |2 t+1|}{2}-\frac{3 \log |3 t+2|}{3}$$ + C
= log |2 log x + 1| – log |3 log x + 2| + C

Question 18.
(i) ∫ $$\frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}$$
(ii) ∫ $$\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}$$ dx (NCERT)
Solution:
(i) put x2 = t
∴ $$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{(t+1)(t+4)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+4}$$ …………….(1)
Multiply both sides by (t + 1) (t + 4) ; we have
1 = A (t + 4) + B (t + 1) …………..(2)
putting t = – 4, – 1 successively in eqn. (2) ; we have
1 = – 3B
⇒ B = – $$\frac{1}{3}$$
and A = $$\frac{1}{3}$$
∴ from (1) ; we have
$$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{\frac{1}{3}}{x^2+1}-\frac{\frac{1}{3}}{x^2+4}$$
Thus, $$\int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3} \int \frac{1}{x^2+1^2} d x-\frac{1}{3} \int \frac{1}{x^2+2^2} d x$$
= $$\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}$$ + C
[∵ $$\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]

(ii) put x2 = t
∴ $$\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{t}{(t+1)(t+4)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+4}$$ …………(1)
Multiply both sides of eqn. (1) by (t + 1) (t + 4) ; we have
t = A (t + 4) + B (t + 1)
putting t = – 1, – 4 successively in eqn. (2) ; we have
– 1 = 3A
⇒ A = – $$\frac{1}{3}$$
and – 4 = – 3B
⇒ B = $$\frac{4}{3}$$
∴ from (1) ; we get

Question 19.
(i) ∫ $$\frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}$$ dx
(ii) ∫ $$\frac{x^2}{x^4-x^2-12}$$ dx (NCERT Exemplar)
Solution:
(i) putting x2 = y
Then $$\frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}=\frac{y+1}{(y+4)(y+25)}$$
Let $$\frac{y+1}{(y+4)(y+25)}=\frac{\mathrm{A}}{y+4}+\frac{\mathrm{B}}{y+25}$$ …………….(1)
Then y + 1 = A (y + 25) + B (y + 4) …………(2)
putting y = – 4, – 25 successively in eqn. (2) ; we have
– 3 = 21 A
⇒ A = – 1/7
and – 24 = – 21 B
⇒ B = 8/7
from (1) ; we have
$$\frac{y+1}{(y+4)(y+25)}=\frac{-1 / 7}{y+4}+\frac{8 / 7}{y+25}$$

(ii) Let I = ∫ $$\frac{x^2 d x}{x^4-x^2-12}$$
= ∫ $$\frac{x^2 d x}{\left(x^2+3\right)\left(x^2-4\right)}$$
put x2 = t, we have
$$\frac{x^2}{\left(x^2+3\right)\left(x^2-4\right)}=\frac{t}{(t+3)(t-4)}$$
Let $$\frac{t}{(t+3)(t-4)}=\frac{\mathrm{A}}{t+3}+\frac{\mathrm{B}}{t-4}$$ ……………..(1)
Then t = A (t – 4) + B (t + 3) …………….(2)
putting t = – 3, 4 successively in eqn. (2) ; we have
– 3 = – 7A
⇒ A = $$\frac{3}{7}$$
and 4 = 7B
⇒ B = 4/7
∴ From (1) ; we have

Question 20.
(i) ∫ $$\frac{d x}{x^4-1}$$
(ii) ∫ $$\frac{x^2}{1-x^4}$$ dx (NCERT Exemplar)
(iii) ∫ $$\frac{\cos x}{\left(4+\sin ^2 x\right)\left(5-4 \cos ^2 x\right)}$$ dx
Solution:
(i) Let I = ∫ $$\frac{d x}{x^4-1}$$ ;
put x2 = y
⇒ $$\frac{1}{x^4-1}=\frac{1}{t^2-1}$$
= $$\frac{1}{(t-1)(t+1)}$$
Let  …………..(1)
⇒ 1 = A (t – 1) + B (t + 1) ………….(2)
putting t = 1, – 1 successively in eqn. (2) ; we have
1 = 2B
⇒ B = 1/2
and 1 = – 2A
⇒ A = – 1/2
∴ From (1) ;

(ii) put x2 = t, we have
$$\frac{x^2}{1-x^4}=\frac{t}{1-t^2}=\frac{t}{(1-t)(1+t)}$$
Let $$\frac{t}{(1-t)(1+t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}$$ ……………..(1)
Then t = A (1 + t) + B (1 – t)
putting t = 1, – 1 successively in eqn. (2) ; we have
1 = 2A
⇒ A = 1/2
and – 1 = 2B
⇒ B = – 1/2

(iii) Let I = ∫ $$\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left(5-4 \cos ^2 x\right)}$$
= ∫ $$\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left[5-4\left(1-\sin ^2 x\right)\right]}$$
= ∫ $$\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left(1+4 \sin ^2 x\right)}$$
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}$$
put t2 = y
∴ $$\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{1}{(4+y)(1+4 y)}$$
= $$\frac{\mathrm{A}}{4+y}+\frac{\mathrm{B}}{1+4 y}$$ …………(1)
Multiplying both sides of eqn. (1) by (4 + y) (1 + 4y) ; we have
1 = A (1 + 4y) + B (4 + y) …………..(2)
putting y = – 4 in eqn. (2) ; we have
1 = A (t – 16)
⇒ A = $$-\frac{1}{15}$$
putting y = – $$\frac{1}{4}$$ in eqn. (2) ; we have
1 = B (4 – $$\frac{1}{4}$$)
⇒ 1 = B ($$\frac{15}{4}$$)
⇒ B = $$\frac{4}{15}$$
∴ from (1) ; we have

Question 21.
(i) ∫ $$\frac{\sqrt{\cos x}}{\sin x}$$ dx
(ii) ∫ $$\frac{\sin x}{\sin 4 x}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\sqrt{\cos x}}{\sin x}$$ dx
put $$\sqrt{cos x}$$ = t
⇒ cos x = t2
⇒ – sin x dx = 2t dt
∴ I = ∫ $$\frac{t \times 2 t d t}{-\left(1-t^4\right)}$$
= – 2 ∫ $$\frac{t^2}{\left(1-t^2\right)\left(1+t^2\right)}$$ dt ……………..(1)
put t2 = y
∴ $$\frac{t^2}{\left(1-t^2\right)\left(1+t^2\right)}=\frac{y}{(1-y)(1+y)}$$
= $$\frac{\mathrm{A}}{1-y}+\frac{\mathrm{B}}{1+y}$$ ……………..(2)
Multiply both sides of eqn. (1) by (1 – y) (1 + y) ; we have
∴ y = A (1 + y) + B (1 – y) …………………(3)
putting y = 1, – 1 successively in eqn. (3) ; we have
1 = 2A
A = $$\frac{1}{2}$$ and B = – $$\frac{1}{2}$$
∴ from (2); we get

(ii) Let I = ∫ $$\frac{\sin x}{\sin 4 x}$$ dx
= ∫ $$\frac{\sin x d x}{2 \sin 2 x \cos 2 x}$$ dx
= ∫ $$\frac{d x}{4 \cos x\left(1-2 \sin ^2 x\right)}$$ dx
put sin x = t
⇒ cos x dx = dt = ∫ $$\frac{d t}{4\left(1-t^2\right)\left(1-2 t^2\right)}$$ ………………(1)
put t2 = y
∴ $$\frac{1}{\left(1-t^2\right)\left(1-2 t^2\right)}=\frac{1}{(1-y)(1-2 y)}=\frac{\mathrm{A}}{1-y}+\frac{\mathrm{B}}{1-2 y}$$ ………………(2)
Multiply both sides of eqn. (2) by (1 – y) (1 – 2y) ; we get
1 = A(1 – 2y) + B (1 – y)
putting y = 1, $$\frac{1}{2}$$ successively in eqn. (1) ; we have
1 = – A
⇒ A = – 1 and B = 2
∴ from (2) ; we have
$$\frac{1}{\left(1-t^2\right)\left(1-2 t^2\right)}=\frac{-1}{1-t^2}+\frac{2}{1-2 t^2}$$
∴ from (1) ; we get

Question 22.
(i) ∫ $$\frac{1+x^2}{1+x^4}$$ dx
(ii) ∫ $$\frac{1-x^2}{1+x^4}$$ dx
Solution:
(i) Let I = ∫ $$\frac{1+x^2}{1+x^4}$$ dx
Divide Num. and Deno. by x2 ; we get
∴ I = ∫ $$\frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}$$ dx
Put x – $$\frac{1}{x}$$ = t
(1 + $$\frac{1}{x^2}$$) dx = dt ;
On Squaring ;
x2 + $$\frac{1}{x^2}$$ = t2 + 2
∴ I = ∫ $$\frac{d t}{t^2+2}$$
= ∫ $$\frac{d t}{t^2+(\sqrt{2})^2}$$
[using $$\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}$$ + C]
∴ I = $$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)$$ + C
= $$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)$$ + C

(ii) Let I = ∫ $$\frac{1-x^2}{1+x^4}$$ dx
Divide Num. and Deno. by x2 ; we get

Question 23.
(i) ∫ $$\frac{x^2-1}{x^4+x^2+1}$$ dx
(ii) ∫ $$\frac{x^2+4}{x^4+x^2+16}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x^2-1}{x^4+x^2+1}$$ dx
Divide Num. and Deno. by x2 ; we get
= ∫ $$\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}$$ dx
put x + $$\frac{1}{x}$$ = t
⇒ (1 – $$\frac{1}{x^2}$$) dx = dt
On squaring; we get
(x + $$\frac{1}{x}$$)2 = t2
⇒ x2 + $$\frac{1}{x^2}$$ = t2 – 2

(ii) Let I = ∫ $$\frac{x^2+4}{x^4+x^2+16}$$ dx
Divide Numerator and denominator by x2 ; we have
I = ∫ $$\frac{\left(1+\frac{4}{x^2}\right) d x}{x^2+1+\frac{16}{x^2}}$$
put x – $$\frac{4}{x}$$ = t
⇒ (1 + $$\frac{4}{x^2}$$) dx = dt
On squaring; we have
(x – $$\frac{4}{x}$$)2 = t2
⇒ x2 + $$\frac{16}{x^2}$$ – 8 = t2
⇒ x2 + $$\frac{16}{x^2}$$ = t2 + 8
∴ I = ∫ $$\frac{d t}{t^2+8+1}$$
= $$\frac{d t}{t^2+3^2}$$
= $$\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)$$ + C
= $$\frac{1}{3} \tan ^{-1}\left(\frac{x^2-4}{3 x}\right)$$ + C

Question 24.
(i) ∫ $$\frac{1}{x^4+1}$$ dx
(ii) ∫ $$\frac{x^2}{x^4+1}$$ dx
Solution:
(i) ∫ $$\frac{1}{x^4+1}$$ dx
= $$\frac{1}{2} \int \frac{2 d x}{x^4+1}$$
= $$\frac{1}{2} \int \frac{\left[\left(x^2+1\right)-\left(x^2-1\right)\right]}{x^4+1}$$ dx

(ii) Let I = ∫ $$\frac{x^2}{x^4+1}$$ dx
∴ I = $$\frac{1}{2} \int \frac{\left(x^2+1\right)+\left(x^2-1\right)}{x^4+1}$$
= $$\frac{1}{2} \int \frac{x^2+1}{x^4+1} d x+\frac{1}{2} \int \frac{x^2-1}{x^4+1} d x$$
= $$\frac{1}{2}$$ I1 – $$\frac{1}{2}$$ I2 ……………..(1)
where I1 = ∫ $$\frac{x^2}{x^4+1}$$ dx
Divide Num. and Deno. by x2 ; we get

Question 25.
(i) ∫ $$\frac{x^2}{x^4+x^2+1}$$ dx
(ii) ∫ $$\frac{x^2-3 x+1}{x^4-x^2+1}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x^2}{x^4+x^2+1}$$ dx

(ii) Let I = ∫ $$\frac{x^2-3 x+1}{x^4-x^2+1}$$ dx
∴ I = $$\int \frac{\left(x^2+1\right) d x}{x^4-x^2+1}-3 \int \frac{x d x}{x^4-x^2+1}$$
⇒ I = I1 – 3 I2 …………….(1)
where I1 = ∫ $$\frac{\left(x^2+1\right) d x}{x^4-x^2+1}$$
Divide Num. and Deno. by x2 ; we have
= ∫ $$\frac{\left(1+\frac{1}{x^2}\right) d x}{x^2-1+\frac{1}{x^2}}$$
put x – $$\frac{1}{x}$$ = t
⇒ (1 + $$\frac{1}{x^2}$$) dx = dt
On squaring ;
(x – $$\frac{1}{x}$$)2 = t2
⇒ x2 + $$\frac{1}{x^2}$$ – 2 = t2
⇒ x2 + $$\frac{1}{x^2}$$ = t2 + 2

Question 26.
(i) ∫ $$\frac{\sin x+\cos x}{\sin ^2 x+\cos ^4 x}$$ dx
(ii) ∫ ($$\sqrt{\cot x}+\frac{1}{\sqrt{\cot x}}$$) dx
Solution:
(i) Let

put sin x – cos x = t
⇒ (cos x + sin x) dx = dt
On squaring ; we have
(sin x – cos x)2 = t2
⇒ sin2 x + cos2 x – sin2x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2
∴ I = 4 ∫ $$\frac{d t}{4-\left(1-t^2\right)^2}$$
= ∫ $$\frac{4 d t}{\left[2-\left(1-t^2\right)\right]\left[2+1-t^2\right]}$$
= 4 ∫ $$\frac{d t}{\left(1+t^2\right)\left(3-t^2\right)}$$ ……………….(1)
put t2 = y
∴ $$\frac{1}{\left(1+t^2\right)\left(3-t^2\right)}=\frac{1}{(1+y)(3-y)}=\frac{\mathrm{A}}{1+y}+\frac{\mathrm{B}}{3-y}$$ ……………(2)
Multiplying both sides by (1 + y) (3 – y) ; we have
1 = A (3 – y) + B (1 + y) …………(3)
putting y = – 1, 3 successively in eqn. (3) ; we have
1 = 4A
⇒ A = $$\frac{1}{4}$$
and B = $$\frac{1}{4}$$
∴ from (2) ; we have

(ii) Let I = ∫ ($$\sqrt{\cot x}+\frac{1}{\sqrt{\cot x}}$$) dx
= ∫ $\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}$ dx
put $$\sqrt{\tan x}$$ = t
⇒ tan x = t2
sec2 dx = 2t dt
⇒ dx = $$\frac{2 t d t}{1+\tan ^2 x}=\frac{2 t d t}{1+t^4}$$
= ∫ $$\left(t+\frac{1}{t}\right) \frac{2 t d t}{1+t^4}$$
= ∫ $$\frac{2\left(t^2+1\right) d t}{t^4+1}$$
Divide Numerator and denominator by t2 ; we have
= 2 ∫ $$\frac{\left(1+\frac{1}{t^2}\right) d t}{t^2+\frac{1}{t^2}}$$
put t – $$\frac{1}{t}$$ = u
⇒ (1 + $$\frac{1}{t^2}$$) dt = du
On squaring; we have

Question 26 (0ld).
(ii) ∫ $$\sqrt{cot x}$$ dx
Solution:
Let I = ∫ $$\sqrt{cot x}$$ dx
put $$\sqrt{cot x}$$ = t
⇒ cot x = t2
⇒ – cosec2 x dx = 2t dt