Evaluate ∫ 3 x - 1 ( x - 2 ) 2 d x

We have to find ∫ 3 x - 1 x - 2 2 d x ⇒ 3 x - 1 x - 2 2 = A x - 2 + B x - 2 2 ⇒ 3 x - 1 = A x - 2 + B ⇒ 3 x - 1 = A x - 2 + B Equating the coefficients of x and constant terms, we get ⇒ A = 3 and - 2 A + B = - 1 ⇒ A = 3 , B = 5 ∴ 3 x - 1 x - 2 2 = 3 x - 2 + 5 x - 2 2 ⇒ ∫ 3 x - 1 x - 2 2 d x = 3 ∫ d x x - 2 + 5 ∫ d x x - 2 2 = 3 log x - 2 - 5 x - 2 + c ∵ ∫ x - n d x = x - n + 1 - n + 1 + c , ∫ d x x = log x = log x - 2 3 - 5 x - 2 + c ∵ n log m = log m n , log m - log n = log m n Hence, ∫ 3 x - 1 x - 2 2 d x = log x - 2 3 - 5 x - 2 + c

Evaluate ∫ x x - 1 ( x 2 + 1 ) d x

We have to find ∫ x x - 1 x 2 + 1 d x . ⇒ x x - 1 x 2 + 1 = A x + B x 2 + 1 + C x - 1 ⇒ x = A x + B x - 1 + C x 2 + 1 ⇒ x = A x 2 + B - A x - B + C x 2 + 1 Equating the coefficients of x and constant terms, we get ⇒ A + C = 0 , B - A = 1 and C - B = 0 ⇒ A = - 1 2 , B = 1 2 , C = 1 2 ∴ x x - 1 x 2 + 1 = - x 2 + 1 2 x 2 + 1 + 1 1 · 1 x - 1 = - 1 2 · x - 1 x 2 + 1 + 1 2 · 1 x - 1 = - 1 2 · x x 2 + 1 + 1 2 · 1 x 2 + 1 + 1 2 · 1 x - 1 Integrating w.r.t. x I = - 1 2 ∫ x x 2 + 1 d x + 1 2 ∫ d x x 2 + 1 + 1 2 ∫ d x x - 1 ↓ ↓ ↓ I 1 I 2 I 3 Solving I 1 I 1 = - 1 2 ∫ x x 2 + 1 d x , Put x 2 + 1 = t Differentiating both sides w.r.t. x 2 x + 0 = d t d x ⇒ d t 2 x = d x Hence we can write - 1 2 ∫ x x 2 + 1 d x = - 1 2 ∫ x t d t 2 x = - 1 2 × 2 ∫ d t t = - 1 4 log t + c 1 = - 1 4 log x 2 + 1 + c 1 ∵ t = x 2 + 1 Solving I 2 I 2 = 1 2 ∫ d x x 2 + 1 = 1 2 tan - 1 x 1 + c 2 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Hence I = I 1 + I 2 + I 3 ⇒ ∫ x x - 1 x 2 + 2 d x = - 1 4 log x 2 + 1 + c 1 + 1 2 tan - 1 x + 1 2 ∫ d x x - 1 = 1 2 log x - 1 + 1 2 tan - 1 x - 1 4 log x 2 + 1 + c ∵ ∫ d x x = log x

Evaluate ∫ 1 x 4 + 1 d x

We have to find ∫ 1 x 4 + 1 d x . Let I = ∫ 1 x 4 + 1 d x , Divide by x 2 I = ∫ 1 x 2 x 2 + 1 x 2 d x = 1 2 ∫ 2 x 2 x 2 + 1 x 2 d x = 1 2 ∫ 1 + 1 x 2 - 1 - 1 x 2 x 2 + 1 x 2 d x = I = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 d x ⏟ I 1 - 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 d x ⏟ I 2 Solving I 1 I 1 = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 d x = 1 2 ∫ 1 + 1 x 2 x - 1 x 2 + 2 d x Put x - 1 x = t Differentiate w.r.t. x 1 + 1 x 2 = d t d x ⇒ d x = d t 1 + 1 x 2 I 1 = 1 2 ∫ 1 + 1 x 2 t 2 + 2 d t 1 + 1 x 2 = 1 2 ∫ d t t 2 + 2 I = 1 2 2 tan - 1 t 2 + c 1 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put t = x - 1 x Hence, I 1 = 1 2 2 tan - 1 x - 1 x 2 + c 1 = 1 2 2 tan - 1 x 2 - 1 2 x + c 1 Solving I 2 I 2 = 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 d x = 1 2 ∫ 1 - 1 x 2 x + 1 x 2 - 2 d x Put x + 1 x = Y Differentiate w.r.t. x 1 - 1 x 2 = d Y d x ⇒ d x = d Y 1 - 1 x 2 I 2 = 1 2 ∫ 1 - 1 x 2 Y 2 - 2 d Y 1 - 1 x 2 = 1 2 ∫ d Y Y 2 - 2 I 2 = 1 4 2 log Y - 2 Y + 2 + c 2 ∵ ∫ d x x 2 - a 2 = 1 2 a log x - a x + a + c Put Y = x + 1 x Hence, I 2 = 1 4 2 log x + 1 x - 2 x + 1 x + 2 + c 2 = 1 4 2 log x 2 - 2 x + 1 x 2 + 2 x + 1 + c 2 ⇒ I = I 1 - I 2 ⇒ I = 1 2 2 tan - 1 x 2 - 1 2 x - 1 4 2 log x 2 - 2 x + 1 x 2 + 2 x + 1 + c , where c = c 1 + c 2

Evaluate ∫ x 2 x 4 + 1 d x

We have to find ∫ x 2 x 4 + 1 d x . Let I = ∫ x 2 x 4 + 1 d x = 1 2 ∫ 2 x 2 1 + x 4 d x = 1 2 ∫ x 2 + 1 + x 2 - 1 x 4 + 1 d x , Divide by x 2 I = 1 2 ∫ 1 + 1 x 2 + 1 - 1 x 2 x 2 + 1 x 2 d x I = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 d x ⏟ I 1 + 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 d x ⏟ I 2 Solving I 1 I 1 = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 d x = 1 2 ∫ 1 + 1 x 2 x - 1 x 2 + 2 d x Put x - 1 x = t Differentiate w.r.t. x 1 + 1 x 2 = d t d x ⇒ d x = d t 1 + 1 x 2 I 1 = 1 2 ∫ 1 + 1 x 2 t 2 + 2 d t 1 + 1 x 2 = 1 2 ∫ d t t 2 + 2 I = 1 2 2 tan - 1 t 2 + c 1 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put t = x - 1 x Hence, I 1 = 1 2 2 tan - 1 x - 1 x 2 + c 1 = 1 2 2 tan - 1 x 2 - 1 2 x + c 1 Solving I 2 I 2 = 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 d x = 1 2 ∫ 1 - 1 x 2 x + 1 x 2 - 2 d x Put x + 1 x = Y Differentiate w.r.t. x 1 - 1 x 2 = d Y d x ⇒ d x = d Y 1 - 1 x 2 I 2 = 1 2 ∫ 1 - 1 x 2 Y 2 - 2 d Y 1 - 1 x 2 = 1 2 ∫ d Y Y 2 - 2 I 2 = 1 4 2 log Y - 2 Y + 2 + c 2 ∵ ∫ d x x 2 - a 2 = 1 2 a log x - a x + a + c Put Y = x + 1 x Hence, I 2 = 1 4 2 log x + 1 x - 2 x + 1 x + 2 + c 2 = 1 4 2 log x 2 - 2 x + 1 x 2 + 2 x + 1 + c 2 ⇒ I = I 1 + I 2 ⇒ I = 1 2 2 tan - 1 x 2 - 1 2 x + 1 4 2 log x 2 - 2 x + 1 x 2 + 2 x + 1 + c , where c = c 1 + c 2 .

Evaluate ∫ x 2 x 4 + x 2 + 1 d x

We have to find ∫ x 2 x 4 + x 2 + 1 d x . Let I = ∫ x 2 x 4 + x 2 + 1 d x = 1 2 ∫ 2 x 2 x 4 + x 2 + 1 d x = 1 2 ∫ x 2 + 1 + x 2 - 1 x 4 + x 2 + 1 d x , Divide by x 2 I = 1 2 ∫ 1 + 1 x 2 + 1 - 1 x 2 x 2 + 1 x 2 + 1 d x I = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 + 1 d x ⏟ I 1 + 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 + 1 d x ⏟ I 2 Solving I 1 I 1 = 1 2 ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 + 1 d x = 1 2 ∫ 1 + 1 x 2 x - 1 x 2 + 3 d x Put x - 1 x = t Differentiate w.r.t. x 1 + 1 x 2 = d t d x ⇒ d x = d t 1 + 1 x 2 I 1 = 1 2 ∫ 1 + 1 x 2 t 2 + 3 d t 1 + 1 x 2 = 1 2 ∫ d t t 2 + 3 I = 1 2 3 tan - 1 t 3 + c 1 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put t = x - 1 x Hence, I 1 = 1 2 3 tan - 1 x - 1 x 3 + c 1 = 1 2 3 tan - 1 x 2 - 1 3 x + c 1 Solving I 2 I 2 = 1 2 ∫ 1 - 1 x 2 x 2 + 1 x 2 + 2 - 2 + 1 d x = 1 2 ∫ 1 - 1 x 2 x + 1 x 2 - 1 d x Put x + 1 x = Y Differentiate w.r.t. x 1 - 1 x 2 = d Y d x ⇒ d x = d Y 1 - 1 x 2 I 2 = 1 2 ∫ 1 - 1 x 2 Y 2 - 1 d Y 1 - 1 x 2 = 1 2 ∫ d Y Y 2 - 1 I 2 = 1 4 log Y - 1 Y + 1 + c 2 ∵ ∫ d x x 2 - a 2 = 1 2 a log x - a x + a + c Put Y = x + 1 x Hence, I 2 = 1 4 log x + 1 x - 1 x + 1 x + 1 + c 2 = 1 4 log x 2 - x + 1 x 2 + x + 1 + c 2 ⇒ I = I 1 + I 2 ⇒ I = 1 2 3 tan - 1 x 2 - 1 3 x + 1 4 log x 2 - x + 1 x 2 + x + 1 + c , where c = c 1 + c 2 .

Evaluate ∫ x 2 - 3 x + 1 x 4 - x 2 + 1 d x

We have to find ∫ x 2 - 3 x + 1 x 4 - x 2 + 1 d x . Let I = ∫ x 2 - 3 x + 1 x 4 - x 2 + 1 d x = ∫ x 2 + 1 - 3 x x 4 - x 2 + 1 d x , Divide by x 2 I = ∫ 1 + 1 x 2 - 3 x x 2 + 1 x 2 - 1 d x I = ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 - 1 d x ⏟ I 1 - ∫ 3 x x 4 - x 2 + 1 d x ⏟ I 2 Solving I 1 I 1 = ∫ 1 + 1 x 2 x 2 + 1 x 2 - 2 + 2 - 1 d x = ∫ 1 + 1 x 2 x - 1 x 2 + 1 d x Put x - 1 x = t Differentiate w.r.t. x 1 + 1 x 2 = d t d x ⇒ d x = d t 1 + 1 x 2 I 1 = ∫ 1 + 1 x 2 t 2 + 1 d t 1 + 1 x 2 = ∫ d t t 2 + 1 I = tan - 1 t + c 1 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put t = x - 1 x Hence, I 1 = tan - 1 x - 1 x + c 1 = tan - 1 x 2 - 1 x + c 1 Solving I 2 I 2 = ∫ 3 x x 4 - x 2 + 1 d x Put x 2 = Y Differentiate w.r.t. x 2 x = d Y d x ⇒ d x = d Y 2 x I 2 = ∫ 3 x Y 2 - Y + 1 d Y 2 x = 3 2 ∫ d Y Y 2 - Y + 1 4 - 1 4 + 1 = 3 2 ∫ d Y Y - 1 2 2 + 3 4 I 2 = 3 tan - 1 2 Y - 1 2 3 + c 2 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put Y = x 2 Hence, I 2 = 3 tan - 1 2 x 2 - 1 3 + c 2 ⇒ I = I 1 - I 2 ⇒ I = tan - 1 x 2 - 1 x - 3 tan - 1 2 x 2 - 1 3 + c , where c = c 1 + c 2 .

Evaluate ∫ sin x + cos x sin 2 x + cos 4 x d x

Let I = ∫ sin x + cos x sin 2 x + cos 4 x d x Substitute sin 2 x = 1 - cos 2 x 2 , cos 2 x = 1 + cos 2 x 2 Therefore, I = ∫ sin x + cos x d x 1 4 ( 2 - 2 cos 2 x + 1 + 2 cos 2 2 x + 2 cos 2 x ) I = 4 ∫ sin x + cos x d x 3 + cos 2 2 x = 4 ∫ sin x + cos x d x 3 + 1 - sin 2 2 x = 4 ∫ sin x + cos x d x 4 - sin 2 2 x Substitute ( sin x - cos x ) = t Differentiate with respect to x , then ( cos x + sin x ) d x = d t I = 4 ∫ d t 4 - ( 1 - t 2 ) 2 = 4 ∫ d t 4 - 1 - t 4 + 2 t 2 = 4 ∫ d t 3 - t 4 + 2 t 2 = 4 ∫ d t ( 1 + t 2 ) ( 3 - t 2 ) I = ∫ 1 + t 2 + ( 3 - t 2 ) d t ( 1 + t 2 ) ( 3 - t 2 ) I = ∫ d t ( 1 + t 2 ) + ∫ d t ( 3 - t 2 ) I = tan - 1 t + 1 2 3 log | 3 + t 3 - t | + c I = tan - 1 ( sin x - cos x ) + 1 2 3 log | 3 + sin x - cos x 3 - sin x + cos x | + c

Evaluate ∫ cot x d x

We have to find ∫ cot x d x . Let I = ∫ cot x d x Put cot x = t ⇒ t 2 = cot x Differentiating w.r.t. x - cosec 2 x = 2 t d t d x ⇒ d x = 2 t d t - cosec 2 x I = ∫ t · 2 t d t - cosec 2 x = - 2 ∫ t 2 1 + cot 2 x d t = - 2 ∫ t 2 1 + t 4 d t Thus, I = - 2 ∫ t 2 t 4 + 1 d t = - ∫ 2 t 2 1 + t 4 d t = - ∫ t 2 + 1 + t 2 - 1 t 4 + 1 d t , Divide by t 2 I = - ∫ 1 + 1 t 2 + 1 - 1 t 2 t 2 + 1 t 2 d t I = - ∫ 1 + 1 t 2 t 2 + 1 t 2 - 2 + 2 d t ⏟ I 1 - ∫ 1 - 1 t 2 t 2 + 1 t 2 + 2 - 2 d t ⏟ I 2 Solving I 1 I 1 = - ∫ 1 + 1 t 2 t 2 + 1 t 2 - 2 + 2 d t = - ∫ 1 + 1 t 2 t - 1 t 2 + 2 d t Put t - 1 t = A Differentiate w.r.t. t 1 + 1 t 2 = d A d t ⇒ d t = d A 1 + 1 t 2 I 1 = - ∫ 1 + 1 t 2 A 2 + 2 d A 1 + 1 t 2 = - ∫ d A A 2 + 2 I = - 1 2 tan - 1 A 2 + c 1 ∵ ∫ d x x 2 + a 2 = 1 a tan - 1 x a + c Put A = t - 1 t Hence, I 1 = - 1 2 tan - 1 t - 1 t 2 + c 1 = - 1 2 tan - 1 t 2 - 1 2 t + c 1 Solving I 2 I 2 = - ∫ 1 - 1 t 2 t 2 + 1 t 2 + 2 - 2 d t = - ∫ 1 - 1 t 2 t + 1 t 2 - 2 d t Put t + 1 t = Y Differentiate w.r.t. t 1 - 1 t 2 = d Y d t ⇒ d t = d Y 1 - 1 t 2 I 2 = - ∫ 1 - 1 t 2 Y 2 - 2 d Y 1 - 1 t 2 = - ∫ d Y Y 2 - 2 I 2 = - 1 2 2 log Y - 2 Y + 2 + c 2 ∵ ∫ d x x 2 - a 2 = 1 2 a log x - a x + a + c Put Y = t + 1 t Hence, I 2 = - 1 2 2 log t + 1 t - 2 t + 1 t + 2 + c 2 = - 1 2 2 log t 2 - 2 t + 1 t 2 + 2 t + 1 + c 2 ⇒ I = I 1 + I 2 ⇒ I = - 1 2 tan - 1 cot x - 1 2 cot x - 1 2 2 log cot x - 2 cot x + 1 cot x + 2 cot x + 1 + c where t = cot x and c = c 1 + c 2

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 8 - Integrals>EXERCISE

M L Aggarwal Solutions for Chapter: Integrals, Exercise 1: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Integrals, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 8: Integrals, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.