OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(b)

Regular engagement with ISC Class 12 Maths Solutions OP Malhotra Chapter 15 Indefinite Integral-3 Ex 15(b) can boost students’ confidence in the subject.

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(b)

Question 1.
\(\int \frac{x-1}{(x+1)(x-2)} d x\)
Solution:
Let \(\int \frac{x-1}{(x+1)(x-2)} d x\) = \(\frac{\mathrm{A}}{x+1}\) + \(\frac{\mathrm{B}}{x-2}\) …(1)
Multiplying both sides of eqn. (1) by (x + 1)(x – 2); we have
(x – 1) = A(x – 2) + B(x + 1) …(2)
putting x = -1, 2 successively in eqn. (2); we have
-2 = -3 A ⇒ A = 2/3
& 1 = 3B ⇒ B = 1/3
∴ from(1); we have
\(\int \frac{x-1}{(x+1)(x-1)} d x\) = \(\frac{2}{3}\)\(\int \frac{1}{x+1} d x\)+\(\frac{1}{3}\)\(\int \frac{d x}{x-2}\) = \(\frac{2}{3}\)log|x + 1|+ \(\frac{1}{3}\)log|x – 2| + C

Question 2.
\(\int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x\)
Solution:
Let \(\frac{2 x-1}{(x-1)(x+2)(x-3)} d x\) = \(\frac{\mathrm{A}}{x-1}\)+\(\frac{B}{x+2}\)+\(\frac{C}{x-3}\) …(1)
Multiplying eqn. (1) both sides by (x – 1)(x + 2)(x – 3); we have
2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2) …(2)
putting x = 1, -2, 3 successively in eqn. (1); we have
1 = A(3)(-2) ⇒ A = -1/6
-5 =B(-3)(-5) ⇒ B = -1/3
5 = C(2)5 ⇒ C = 1/2
∴ from (1); we have
I = \(\int \frac{-1 / 6}{x-1} d x\) + \(\int \frac{-1 / 3}{x+2} d x\) + \(\int \frac{1 / 2}{x-3} d x\) = –\(\frac { 1 }{ 6 }\)log|x – 1|-\(\frac { 1 }{ 3 }\)log|x+2|+\(\frac { 1 }{ 2 }\)log|x-3|+C

Question 3.
\(\int \frac{x+7}{x^2+2 x-8} d x\)
Solution:
Let I = \(\int \frac{x+7}{x^2+2 x-8} d x\) = \(\int \frac{x+7}{(x-2)(x+4)} d x\)
Let \(\frac{x+7}{(x-2)(x+4)}\) = \(\frac{\mathrm{A}}{x-2}\) + \(\frac{\mathrm{B}}{x+4}\) …(1)
Multiplying eqn. (1) both sides of by (x – 2)(x + 4); we have
x + 7 = A(x + 4) + B(x – 2) …(2)
putting x = 2 in eqn. (2); we have
9 = 6A ⇒ A = 3/2
putting x = -4 in eqn. (2); we have
3 = -6 ⇒ B = B = -1/2
∴ from (1); we have
I = \(\int \frac{3 / 2}{x-2} d x\) + \(\int \frac{-1 / 2}{x+4} d x\) = \(\frac { 3 }{ 2 }\)log|x-2|-\(\frac { 1 }{ 2 }\)log|x+4|+C

Question 4.
\(\int \frac{x d x}{x^2-3 x+2}\)
Solution:
Let I = \(\int \frac{x d x}{x^2-3 x+2}\) = \(\int \frac{x d x}{(x-1)(x-2)}\)
Let \(\frac{x}{(x-1)(x-2)}\) = \(\frac{\mathrm{A}}{x-1}\)+\(\frac{\mathrm{B}}{x-2}\) …(1)
Multiplying eqn. (1) both sides by
(x – 1)(x – 2); we have
x = A(x – 2) + B(x – 1) …(2)
putting x = 1, 2 successively in eqn. (2); we have
1 = -A ⇒ A = -1 & 2 = B
∴ from (1); we have
Thus, I = \(\int \frac{-1}{x-1} d x\)+\(\int \frac{2}{x-2} d x\) = -log|x – 1|+2 log|x – 2| + C

Question 5.
\(\int \frac{2 x+7}{x^2-x-2} d x\)
Solution:
Let I = \(\int \frac{2 x+7}{x^2-x-2} d x\)
Let \(\frac{2 x+7}{x^2-x-2}\) = \(\frac{2 x+7}{(x+1)(x-2)}\)
= \(\frac{A}{x+1}\) + \(\frac{B}{x-2}\) ….(1)
Multiply both sides of eqn. (1) by (x + 1)(x – 2); we have
2x +7 = A(x – 2) + B(x + 1) …(2)
putting x = 2,-1 successively in eqn. (2); we have
11 = 3 B ⇒ B = \(\frac{11}{3}\)
and 5 = – 3A ⇒ A = –\(\frac{5}{3}\)
∴ from (1); we have
I = –\(\frac{5}{3}\) \(\int \frac{d x}{x+1}\) + \(\frac{11}{3}\)\(\int \frac{d x}{x-2}\)
= –\(\frac{5}{3}\) log|x + 1| + \(\frac{11}{3}\) log| x – 2| + C

Question 6.
\(\int \frac{x+1}{x^2+4 x-5} d x\)
Solution:
Let I = \(\int \frac{x+1}{x^2+4 x-5} d x\) = \(\int \frac{(x+1) d x}{(x-1)(x+5)}\)
Let \(\frac{x+1}{(x-1)(x+5)}\) = \(\frac{\mathrm{A}}{x-1}\) + \(\frac{\mathrm{B}}{x+5}\) …(1)
Multiplying eqn. (1) both sides of by (x – 1)(x + 5); we have
x + 1 = A(x + 5) + B(x – 1) …(2)
putting x = 1, in eqn. (2); we have
2 = 6A ⇒ A = 1/3
putting x = -5 in eqn. (2); we have
-4 = -6 B ⇒ B= 2/3
∴ from (1); we have
I = \(\int \frac{1 / 3}{x-1} d x\) + \(\int \frac{2 / 3}{x+5} d x\)
= \(\frac{1}{3}\)log|x – 1|+\(\frac{2}{3}\)log|x + 5| + C

Question 7.
\(\int \frac{x^2+2 x+8}{(x-1)(x-2)} d x\)
Solution:
Here degree of numerator of integand is equal to degree of denominator so it is an improper fraction.
Let I = \(\frac{x^2+2 x+8}{(x-1)(x-2)}\) = 1 + \(\frac{\mathrm{A}}{x-1}\) + \(\frac{\mathrm{B}}{x-2}\) …(1)
Multiplying both sides of eqn. (1) by (x – 1)(x – 2); we have
x² + 2x + 8 = (x – 1)(x – 2) + A(x – 2) + B(x – 1) …(2)
putting x = 1 in eqn. (2); we have
11 = -A ⇒ A = -11
putting x = 1 in eqn. (2); we have
16 = B
∴ from (1); we have
\(\int \frac{x^2+2 x+8}{(x-1)(x-2)} d x\) = \(\int 1 d x-11 \int \frac{d x}{x-1}\) + \(16 \int \frac{d x}{x-2}\) = x-11 log|x – 1|+ 16 log|x – 2| + C

Question 8.
\(\int \frac{x^2-x-2}{1-x^2} d x\)
Solution:
Since degree of numerator of integrand is equal to degree of denominator then by actual division,

Question 9.
\(\int \frac{x^2+x+1}{(x-1)^3} d x\)
Solution:
Let I = \(\int \frac{x^2+x+1}{(x-1)^3} d x\)
Let \(\frac{x^2+x+1}{(1-x)^3}\) = \(\frac{\mathrm{A}}{x-1}\) + \(\frac{\mathrm{B}}{(x-1)^2}\)+ \(\frac{C}{(x-1)^3}\) …(1)
Multiplying both sides of eqn. (1) by (x – 1)³; we have
x² + x + 1 = A(x – 1)² + B(x – 1) + C
putting x = 1 in eqn. (2); we have 3 = C
Coeff of x²; 1 = A
Coeff of x ; 1 = -2A + B ⇒ B = 3
∴ from (1); we have
I = \(\int \frac{d x}{x-1}\) + \(\int \frac{3}{(x-1)^2} d x\) + \(\int \frac{3}{(x-1)^3} d x\) = log|x – 1|+ 3\(\frac{(x-1)^{-2+1}}{(-2+1)}\) + 3\(\frac{(x-1)^{-3+1}}{-3+1}\)
= log|x – 1| – \(\frac{3}{x-1}\) – \(\frac { 3 }{ 2 }\)\(\frac{1}{(x-1)^2}\) + C

Question 10.
\(\int \frac{\sin \theta \cos \theta}{\cos ^2 \theta-\cos \theta-2} d \theta\)
Solution:
Let I = \(\int \frac{\sin \theta \cos \theta}{\cos ^2 \theta-\cos \theta-2} d \theta\); Put cos θ = t ⇒ – sin θ dθ = dt
∴ I = \(\int \frac{-t d t}{t^2-t-2}\) = \(\int \frac{-t d t}{(t-2)(t+1)}\)
Let \(\frac{-t}{(t-2)(t+1)}\) = \(\frac{\mathrm{A}}{t-2}\) + \(\frac{\mathrm{B}}{t+1}\) ….(1)
Multiplying eqn. (1) both sides of by
(t – 2)(t + 1); we have -t = A(t + 1) + B(t – 2)
putting t = 2, in eqn. (2); we have -2 = 3 A ⇒ A = -2/3
putting t = -1 in eqn. (2); we have
1 = B(-3) ⇒ B = -1/3
∴ from (1); we have

Question 11.
\(\int \frac{3 x-1}{(x-2)^2} d x\)
Solution:
Let \(\frac{3 x-1}{(x-2)^2}\) = \(\frac{A}{x-2}\) + \(\frac{B}{(x-2)^2}\) …(1)
Multiplying both sides of eqn. (1) by (x – 2)²; we have
3x – 1 = A(x – 2) + B …(2)
putting x = 2 in eqn. (2); we have 5 = B
Coeff of x ; 3 = A
∴ from (1); we have
\(\frac{3 x-1}{(x-2)^2}\) = \(\frac{3}{x-2}\) + \(\frac{5}{(x-2)^2}\)
∴ \(\int \frac{3 x-1}{(x-2)^2} d x\) = \(3 \int \frac{1}{x-2} d x+5 \int \frac{d x}{(x-2)^2}\)
= 3 log|x – 2| – \(\frac{5}{x-2}\) + C

Question 12.
(i) \(\int \frac{x^2+x+1}{x^2(x+1)} d x\)
(ii) \(\int \frac{2}{(1-x)\left(1+x^2\right)}\)
Solution:
(i) Let \(\frac{x^2+x+1}{x^2(x+1)}\) = \(\frac{\mathrm{A}}{x}\) + \(\frac{\mathrm{B}}{x^2}\) + \(\frac{C}{x+1}\) …(1)
Multiplying both sides of eqn. (1) by x²(x + 1); we have
x² + x + 1 = Ax(x + 1) + B(x + 1) + C x² …(2)
putting x = 0 in eqn. (2); we have 1 = B
putting x = -1 in eqn. (2); we have 1 = C
Coeff of x²; 1 = A + C ⇒ A = 0
∴ from (1); we have

(ii) Let \(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{\mathrm{A}}{1-x}\) + \(\frac{\mathrm{B} x+\mathrm{C}}{\left(1+x^2\right)}\) …(1)
Multiplying both sides of eqn. (1) by (1 – x) (1 + x²); we get
2 = A(1 + x²) + (B x + C)(1 – x) …(2)
putting x = 1 in eqn. (2); we have 2 = 2 A ⇒ A = 1
Coeff of x²; 0 = A – B ⇒ B = 1; Coeff of x ; 0 = + B – C ⇒ C = 1
∴ from (1); we have
\(\frac{2}{(1-x)\left(1+x^2\right)}\) = \(\frac{1}{1-x}\) + \(\frac{x+1}{1+x^2}\)
⇒ \(\int \frac{2}{(1-x)\left(1+x^2\right)} d x\) = \(\int \frac{1}{1-x} d x\) + \(\)
= \(\frac{\log |1-x|}{-1}\) + \(\frac { 1 }{ 2 }\)\(\int \frac{2 x d x}{1+x^2}\) + \(\int \frac{d x}{1+x^2}\) = -log|1 – x|\(\frac { 1 }{ 2 }\)log(1 + x²)+tan^-1x + C
[∵\(\int \frac{f^{\prime}(x) d x}{f(x)}\) = log| \(f(x) \mid \& \int \frac{d x}{x^2+a^2}\) = \(\frac { 1 }{ a }\)tan^-1\(\frac { x }{ a }\)]

Question 13.
\(\int \frac{3 x-2}{(x+1)^2(x+3)} d x\)
Solution:
Let \(\frac{3 x-2}{(x+1)^2(x+3)} d x\) = \(\frac{\mathrm{A}}{x+1}\) + \(\frac{\mathrm{B}}{(x+1)^2}\) + \(\frac{C}{x+3}\) …(1)
Multiplying both sides of eqn. (1) by (x + 1)²(x + 3); we get
3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)² …(2)
putting x = -1 in eqn. (2); we have
– 5 = 2B ⇒ B = -5/2
putting x = – 3 in eqn. (2); we have
– 11 = 4C ⇒ C = \(\frac { -11 }{ 4 }\)
Coeff of x² ; 0 = A + C ⇒ A = \(\frac { 11 }{ 4 }\)
∴ from (1); we have

Question 14.
\(\int \frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)}\)
Solution:
Let I = \(\int \frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)}\); Put x² = t ⇒ 2x dx = dt
∴ I = \(\int \frac{d t}{(t+1)(t+2)}\)
Let \(\frac{1}{(t+1)(t+2)}\) = \(\frac{\mathrm{A}}{t+1}\) + \(\frac{\mathrm{B}}{t+2}\) …(1)
Multiplying both sides of eqn. (1) by (t + 1)(t + 2); we have
I = A(t + 2) + B(t + 1) …(2)
putting t = – 1 in eqn. (2); we have
1 = – A
putting t = – 2 in eqn. (2); we have
1 = – B ⇒ B = – 1
∴ from (1); we have
I = \(\int \frac{1}{t+1} d t\) – \(\int \frac{1}{t+2}\) = log |t + 1|- log |t + 2| + C
Thus I = log\(\left|\frac{t+1}{t+2}\right|\) + C = log\(\left|\frac{x^2+1}{x^2+2}\right|\) + C

Question 15.
\(\int \frac{x^2+1}{x^2-1} d x\)
Solution:
Let I = \(\int \frac{x^2+1}{x^2-1} d x\)
Here, degree of numerator of integrand is equal to degree of denominator.
Thus, \(\frac{x^2+1}{x^2-1}\) = 1 + \(\frac{\mathrm{A}}{x-1}\) + \(\frac{\mathrm{B}}{(x+1)}\) …(1)
Multiplying both sides of eqn. (1) by (x² – 1); we have
x² + 1 = x² – 1 + A(x + 1) + B(x – 1)
putting x = 1, -1 successively in eqn. (2); we have
2 = 0 + 2A ⇒ A = 1 & 2 = 0 – 2B ⇒ B = -1
∴ from (1); we have
I = \(\int\left[1+\frac{1}{x-1}-\frac{1}{x+1}\right] d x\) = x + log |x – 1|- log |x + 1| + C

Question 16.
\(\int \frac{x^2}{x^4+x^2-2} d x\)
Solution:
Put x = t = \(\frac{x^2}{x^4+x^2-2}\) = \(\frac{t}{t^2+t-2}\) = \(\frac{t}{(t-1)(t+2)}\) …(1)
Let \(\frac{t}{(t-1)(t+2)}\) = \(\frac{\mathrm{A}}{t-1}\) + \(\frac{\mathrm{B}}{t+2}\) …(2)
Multiplying both sides of eqn. (2) by (t – 1)(t + 2); we have
t = A(t + 2) + B(t – 1) …(3)
putting t = 1, – 2 successively in eqn. (3); we have
1 = 3 A ⇒ A = 1/3 & – 2 = – 3B ⇒ B = 2/3
∴ from (1) & (2); we have
\(\frac{x^2}{x^4+x^2-2}\) = \(\frac{1 / 3}{x^2-1}\) + \(\frac{2 / 3}{x^2+2}\)
⇒ \(\int \frac{x^2}{x^4+x^2-2} d x\) = \(\frac { 1 }{ 3 }\)\(\int \frac{d x}{x^2-1}\) + \(\frac { 2 }{ 3 }\)\(\int \frac{d x}{x^2+2}\)
= \(\frac { 1 }{ 3 }\)\(\int \frac{d x}{x^2-1^2}\) + \(\frac { 2 }{ 3 }\)\(\int \frac{d x}{x^2+(\sqrt{2})^2}\) = \(\frac { 1 }{ 3 }\) x \([latex]\frac { 1 }{ 2 }\)[/latex]log\(\left|\frac{x-1}{x+1}\right|\) + \(\frac { 2 }{ 3 }\) × \(\frac{1}{\sqrt{2}}\)tan^-1\(\left(\frac{x}{\sqrt{2}}\right)\) + C
= \(\frac { 1 }{ 6 }\)log_e\(\left|\frac{x-1}{x+1}\right|\) + \(\frac{\sqrt{2}}{3}\)tan^-1\(\left(\frac{x}{\sqrt{2}}\right)\) + C

Question 17.
\(\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x\)
Solution:
Let \(\frac{x^2+x+1}{(x+1)^2(x+2)}\) = \(\frac{\mathrm{A}}{x \times 1}\) + \(\frac{\mathrm{B}}{(x+1)^2}\) + \(\frac{B}{x+2}\) …(1)
Multiplying both sides of eqn. (1) by (x + 1)² (x + 2); we have
x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 2)² …(2)
putting x = -1, -2 successively in eqn. (2); we have
1 = B & 3 = C
Coeff. of x²; 1 = A + C ⇒ A = 1 – C = 1 – 3 = – 2
∴ from (1); we have
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) = \(\frac{-2}{x+1}\) + \(\frac{1}{(x+1)^2}\) + \(\frac{3}{x+2}\)
∴ \(\int \frac{x^2+x+1}{(x+1)^2(x+2)} d x\)
= – 2\(\int \frac{d x}{x+1}\) + \(\int \frac{d x}{(x+1)^2}\) + \(3 \int \frac{d x}{x+2}\) = -2 log |x + 1| + \(\frac{(x+1)^{-2+1}}{-2+1}\) + 3 log|x + 2| + C
= -2 log| x + 1| – \(\frac{1}{x+1}\) + 3 log|x + 2| + C

Question 18.
\(\int \frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)} d x\)
Solution:
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)
Multiplying both sides of eqn. (2) by (t + 1)(t + 4); we have
t = A(t + 4) + B(t +1) …(2)
putting t = -1, -4 successivelyin eqn. (2); we have
-1 = 3A ⇒ A = -1/3 & – 4 = – 3B ⇒ B = 4/3
∴ from (1); we have
\(\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}\) = \(\frac{-1 / 3}{x^2+1}\) + \(\frac{4 / 3}{x^2+4}\)
∴ \(\int \frac{x^2 d x}{\left(x^2+1\right)\left(x^2+4\right)}\) = – \(\frac { 1 }{ 3 }\)\(\int \frac{d x}{x^2+1}\) + \(\frac { 4 }{ 3 }\)\(\int \frac{d x}{x^2+2^2}\)
= – \(\frac { 1 }{ 3 }\)tan^-1x + \(\frac { 4 }{ 3 }\) × \(\frac { 1 }{ 2 }\)tan^-1\(\left(\frac{x}{2}\right)\)+C = –\(\frac { 1 }{ 3 }\)tan^-1 x + \(\frac { 2 }{ 3 }\)tan^-1\(\left(\frac{x}{2}\right)\) + C

Question 19.
\(\int \frac{d x}{1+x+x^2+x^3}\)
Solution:
Let I = \(\int \frac{d x}{1+x+x^2+x^3}\) = \(\int \frac{d x}{(1+x)\left(1+x^2\right)}\)
Let \(\frac{1}{(1+x)\left(1+x^2\right)}\) = \(\frac{\mathrm{A}}{1+x}\) + \(\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}\) …(1)
Multiplying both sides of eqn. (1) by (x + 1)(1 + x²); we have
1 = A(1 + x²) + (Bx + C)(1 + x) ….(2)
putting x = -1 in eqn. (2); we have
1 = 2 A ⇒ A = 1/2
Coeff. of x²; 0 = A + B ⇒ B = -1/2
Coeff. of x; 0 = B + C ⇒ C = + 1/2
∴ from (1); we have

Question 20.
\(\int \frac{x^2 d x}{\left(x^2-1\right)\left(x^2+2\right)}\)
Solution:
Put x² = t
\(\frac{x^2}{\left(x^2-1\right)\left(x^2+2\right)}\) = \(\frac{t}{(t-1)(t+2)}\) = \(\frac{\mathrm{A}}{t-1}\) + \(\frac{\mathrm{B}}{t+2}\) …(1)
Multiplying both sides of eqn. (1) by (t – 1)(t + 2); we have
t = A(t + 2) + B(t – 1) …(2)
putting t = 1, -2 successively in eqn. (2); we have
1 = 3A ⇒ A = 1/3 & – 2 = – 3B ⇒ B = 2/3
∴ from eqn. (1); we have

Question 21.
\(\int \frac{2 x}{x^3-1} d x\)
Solution:
Let I = \(\int \frac{2 x}{x^3-1} d x\) = \(\int \frac{2 x d x}{(x-1)\left(x^2+x+1\right)}\)
Let \(\frac{2 x}{(x-1)\left(x^2+x+1\right)}\) = \(\frac{\mathrm{A}}{x-1}\) + \(\frac{\mathrm{B} x+\mathrm{C}}{x^2+x+1}\) …(1)
Multiplying both sides of eqn. (1) by (x – 1)(x² + x + 1); we have
2x = A(x² + x + 1) + (Bx + C)(x – 1)
putting x = 1 in eqn. (2); we have
2 = 3A ⇒ A = 2/3
Coeff. of x²; 0 = A + B ⇒ B = – 2/3
Coeff. of x ; 2 = A – B + C
⇒ C = 2 – \(\frac { 2 }{ 3 }\) – \(\frac { 2 }{ 3 }\) = \(\frac { 2 }{ 3 }\)
∴ from (1); we have

Question 22.
\(\int \frac{d x}{x+x^2+x^3}\)
Solution:
Let I = \(\int \frac{d x}{x+x^2+x^3}\)
Let \(\frac{1}{x\left(1+x+x^2\right)}\) = \(\frac{\mathrm{A}}{x}\) + \(\frac{\mathrm{B} x+\mathrm{C}}{1+x+x^2}\)
Multiplying both sides of eqn. (1) by x(1 + x + x²); we have
1 = A(1 + x + x²) + (Bx +C) x …(2)
putting x = 0 in eqn. (2); we have
1 = A
Coeff. of x² ; 0 = A + B ⇒ B = -1
Coeff. of x ; 0 = A + C ⇒ C = – 1
∴ from (1); we have

Question 23.
\(\int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} d \theta\)
Solution:
Let I = \(\int \frac{\tan \theta+\tan ^3 \theta}{1+\tan ^3 \theta} d \theta\) = \(\int \frac{\tan \theta\left(1+\tan ^2 \theta\right)}{1+\tan ^3 \theta} d \theta\) = \(\int \frac{\tan \theta \sec ^2 \theta}{1+\tan ^3 \theta} d \theta\)
Put tan θ = t ⇒ sec²θ dθ = dt
∴ I = \(\int \frac{t d t}{1+t^3}\) = \(\int \frac{t d t}{(1+t)\left(t^2-t+1\right)}\)
Let \(\frac{t}{(1+t)\left(t^2-t+1\right)}\) = \(\frac{\mathrm{A}}{1+t}\) + \(\frac{\mathrm{B} t+\mathrm{C}}{t^2-t+1}\) …(1)
Multiplying both sides of eqn. (1) by (1 + t)(t² – t + 1); we have
t = A(t² – t + 1) + (Bt + C)(t + 1) ….(2)
putting t = -1 in eqn. (2); we have
-1 = A(1 + 1 + 1) ⇒ A = -1/3
Coeff. of t²; 0 = A + B ⇒ B = 1/3
Coeff. of t ; 1 = -A + B + C

Question 24.
\(\int \frac{\sec ^2 \theta d \theta}{\tan ^3 \theta+4 \tan \theta}\)
Solution:
Let = \(\int \frac{\sec ^2 \theta d \theta}{\tan ^3 \theta+4 \tan \theta}\); Put tan θ = t ⇒ sec²θ dθ = dt
∴ I = \(\int \frac{d t}{t^3-4 t}\) = \(\int \frac{d t}{t\left(t^2-4\right)}\) …(1)
Let \(\frac{d t}{t\left(t^2-4\right)}\) = \(\frac{\mathrm{A}}{t}\) + \(\frac{\mathrm{B} t+\mathrm{C}}{t^2+4}\)
Multiplying both sides of eqn. (1) byt (t² + 4); we have
1 = A(t² + 4) + (Bt + C) t ….(2)
putting t = 0 in eqn. (2); we have
1 = 4A ⇒ A = 1/4
Coeff. of t²; 0 = A + B
⇒ B = – 1/4
Coeff. of t ; 0 = C
∴ from (1); we have

Question 25.
\(\int \frac{1}{\sin x+\tan x} d x\)
Solution: