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

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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 x^n-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 e^x = t
⇒ e^x 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 e^x = t
⇒ e^x 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) ∫ \(\frac{d x}{\sin x(3+2 \cos x)}\)
Answer:
(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
= ∫ [latex]\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}[/latex] 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