Evaluate the following integral: ∫ 4 x + 1 2 x 2 + x − 7 d x .

Let I = ∫ 4 x + 1 2 x 2 + x − 7 d x Here, d d x ( 2 x 2 + x − 7 ) = 4 x + 1 We know that ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c . Here, f ( x ) = ( 2 x 2 + x - 7 ) ∴ I = ∫ ( 4 x + 1 ) ( 2 x 2 + x − 7 ) - 1 2 d x ⇒ I = ( 2 x 2 + x − 7 ) - 1 2 + 1 - 1 2 + 1 + c ⇒ I = 2 2 x 2 + x − 7 + c Thus, we get ∫ 4 x + 1 2 x 2 + x − 7 d x = 2 2 x 2 + x − 7 + c where c is an integration constant.

Evaluate the following integral: ∫ 3 x + 1 ( 3 x 2 + 2 x − 5 ) 3 d x

Let I = ∫ 3 x + 1 ( 3 x 2 + 2 x − 5 ) 3 d x Here, d d x ( 3 x 2 + 2 x − 5 ) = 6 x + 2 ∴ I = 1 2 ∫ 6 x + 2 ( 3 x 2 + 2 x − 5 ) 3 d x ⇒ I = 1 2 ∫ ( 6 x + 2 ) ( 3 x 2 + 2 x − 5 ) - 3 d x ⇒ I = 1 2 ( 3 x 2 + 2 x − 5 ) - 3 + 1 ( - 3 + 1 ) + c [ ∵ ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c ] ⇒ I = - 1 4 ( 3 x 2 + 2 x − 5 ) 2 + c where c is an integration constant

Evaluate the following integral: ∫ ( 4 x + 2 ) x 2 + x − 3 d x

Let I = ∫ ( 4 x + 2 ) x 2 + x − 3 d x Here, d d x ( x 2 + x − 3 ) = 2 x + 1 ∴ I = ∫ 2 ( 2 x + 1 ) x 2 + x − 3 d x ⇒ I = 2 ∫ ( 2 x + 1 ) x 2 + x − 3 d x ⇒ I = 2 × ( x 2 + x − 3 ) 1 2 + 1 1 2 + 1 + c [ ∵ ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c ] ⇒ ∫ ( 4 x + 2 ) x 2 + x − 3 d x = 4 ( x 2 + x − 3 ) 3 2 3 + c where c is an integration constant.

Evaluate the following integral: ∫ e x ( a + b e x ) 6 d x ; b ≠ 0

Let I = ∫ e x ( a + b e x ) 6 d x where b ≠ 0 Here, d d x ( a + b e x ) = b e x We know that ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c . ∴ I = 1 b ∫ b e x ( a + b e x ) 6 d x ⇒ I = 1 b ( a + b e x ) 6 + 1 6 + 1 + c ⇒ I = ( a + b e x ) 7 7 b + c ⇒ ∫ e x ( a + b e x ) 6 d x = ( a + b e x ) 7 7 b + c where c is an integration constant

Evaluate the following integral: ∫ ( log x ) 2 x d x

Let I = ∫ ( log x ) 2 x d x Here, d d x log x = 1 x We know that ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c ∴ I = ( log x ) 2 + 1 2 + 1 + c ⇒ I = ( log x ) 3 3 + c ⇒ ∫ ( log x ) 2 x d x = ( log x ) 3 3 + c where c is an integration constant.

Evaluate the integral : ∫ sin x ( 1 + cos x ) 2 d x

Let I = ∫ sin x ( 1 + cos x ) 2 d x Put 1 + cos x = t Differentiating both sides, we get ⇒ - sin x d x = d t ∴ I = ∫ - 1 t 2 d t = - t - 2 + 1 - 2 + 1 + c = 1 t + c ∴ I = 1 1 + cos x + c .

Evaluate ∫ tan x sec 2 x 1 - tan 2 x d x .

Let I = ∫ tan x sec 2 x 1 - tan 2 x d x Substitute 1 - tan 2 x = t Differentiating on both sides, we get - 2 tan x · sec 2 x d x = d t ⇒ tan x · sec 2 x d x = d t - 2 ∴ I = ∫ d t - 2 t ⇒ I = - 1 2 log t + c ⇒ I = - 1 2 log 1 - tan 2 x + c .

Evaluate ∫ x 5 1 + x 6 d x .

Let I = ∫ x 5 1 + x 6 d x Substitute 1 + x 6 = t Differentiating on both sides, we get 6 x 5 d x = d t ⇒ x 5 d x = d t 6 ∴ I = ∫ d t 6 t ⇒ I = 1 6 log t + c ⇒ I = 1 6 log 1 + x 6 + c .

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 4: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Integrals, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 8: Integrals, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.