Evaluate ∫ d x 5 - 4 x - 2 x 2

Let I = ∫ d x 5 - 4 x - 2 x 2 I = ∫ d x - 2 x 2 + 2 x - 5 2 = 1 2 ∫ d x - x 2 + 2 x + 1 2 - 1 2 - 5 2 = 1 2 ∫ d x 7 2 - x + 1 2 = 1 2 ∫ d x 7 2 2 - x + 1 2 = 1 2 sin - 1 x + 1 7 2 + c I = 1 2 sin - 1 2 x + 1 7 + c

Evaluate ∫ x - 1 x 2 - x d x

Let I = ∫ x - 1 x 2 - x d x I = ∫ 1 2 2 x - 1 - 1 2 x 2 - x d x I = 1 2 ∫ 2 x - 1 x 2 - x d x - 1 2 ∫ d x x 2 - x For first integral let x 2 - x = t ( 2 x - 1 ) d x = d t I = 1 2 ∫ d t t - 1 2 ∫ d x x 2 - x + 1 2 2 - 1 2 2 I = 1 2 ∫ t - 1 2 d t - 1 2 ∫ d x x - 1 2 2 - 1 2 2 I = 1 2 t 1 2 1 2 - 1 2 log | x - 1 2 + x - 1 2 2 - 1 2 2 I = t - 1 2 log | x - 1 2 + x 2 - x | + c I = x 2 - x - 1 2 log | x - 1 2 + x 2 - x | + c

Evaluate ∫ x + 3 5 - 4 x - x 2 d x

Let I = ∫ x + 3 5 - 4 x - x 2 d x I = ∫ - 1 2 - 4 - 2 x + 1 5 - 4 x - x 2 d x I = - 1 2 ∫ - 4 - 2 x 5 - 4 x - x 2 d x + ∫ 1 5 - 4 x - x 2 d x For first Integral 5 - 4 x - x 2 = t - 4 - 2 x d x = d t I = - 1 2 ∫ d t t + ∫ d x - x 2 + 4 x - 5 I = - 1 2 ∫ t - 1 2 d t + ∫ d x - x 2 + 4 x + 2 2 - 2 2 - 5 I = - 1 2 t 1 2 1 2 + ∫ d x 3 2 - x + 2 2 I = - 5 - 4 x - x 2 + sin - 1 x + 2 3 + c

Evaluate ∫ 5 x + 3 x 2 + 4 x + 10 d x

Let I = ∫ 5 x + 3 x 2 + 4 x + 10 d x I = ∫ 5 2 2 x + 4 - 7 x 2 + 4 x + 10 d x I = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 d x - 7 ∫ 1 x 2 + 4 x + 10 d x For first integral Substitute x 2 + 4 x + 10 = t Differentiate with respect to x , then ( 2 x + 4 ) d x = d t I = 5 2 ∫ d t t - 7 ∫ d x x 2 + 4 x + 2 2 - 2 2 + 10 I = 5 2 ∫ t - 1 2 d t - 7 ∫ d x x + 2 2 + 6 2 I = 5 2 t 1 2 1 2 - 7 log ( x + 2 ) + x + 2 2 + 6 2 + C = 5 x 2 + 4 x + 10 - 7 log x + 2 + x 2 + 4 x + 10 + C

Evaluate ∫ 3 x + 5 x 2 - 8 x + 7 d x

Let I = ∫ 3 x + 5 x 2 - 8 x + 7 d x I = ∫ 3 2 2 x - 8 + 17 x 2 - 8 x + 7 d x I = 3 2 ∫ 2 x - 8 x 2 - 8 x + 7 d x + 17 ∫ 1 x 2 - 8 x + 7 d x Let x 2 - 8 x + 7 = t ( 2 x - 8 ) d x = d t I = 3 2 ∫ d t t + 17 ∫ 1 x 2 - 8 x + 4 2 - 4 2 + 7 d x I = 3 2 ∫ t - 1 2 d t + 17 ∫ d x x - 4 2 - 3 2 I = 3 2 t 1 2 1 2 + 17 log | ( x - 4 ) + x 2 - 8 x + 7 | + c I = 3 x 2 - 8 x + 7 + 17 log | ( x - 4 ) + x 2 - 8 x + 7 | + c

Evaluate ∫ x + 2 ( x - 2 ) ( x - 3 ) d x

Let I = ∫ x + 2 ( x - 2 ) ( x - 3 ) d x I = ∫ x + 2 x 2 - 5 x + 6 d x I = ∫ 1 2 2 x - 5 + 9 2 x 2 - 5 x + 6 d x I = 1 2 ∫ 2 x - 5 x 2 - 5 x + 6 d x + 9 2 ∫ d x x 2 - 5 x + 5 2 2 - 5 2 2 + 6 I = 1 2 ∫ 2 x - 5 x 2 - 5 x + 6 d x + 9 2 ∫ d x x - 5 2 2 - 25 4 + 6 I = 1 2 ∫ 2 x - 5 x 2 - 5 x + 6 d x + 9 2 ∫ d x x - 5 2 2 - 1 4 I = 1 2 ∫ 2 x - 5 x 2 - 5 x + 6 d x + 9 2 ∫ d x x - 5 2 2 - 1 2 2 Using: ∫ d x x 2 - a 2 = log x + x 2 - a 2 + c and ∫ f ' x f x d x = 2 f x + c Hence, I = x 2 - 5 x + 6 + 9 2 log | x - 5 2 + x 2 - 5 x + 6 | + c .

Evaluate ∫ x + 2 x 2 + 2 x + 3 d x

Let I = ∫ x + 2 x 2 + 2 x + 3 d x I = ∫ 1 2 2 x + 2 x 2 + 2 x + 3 d x + ∫ 1 x 2 + 2 x + 3 d x I = I = ∫ 1 2 2 x + 2 x 2 + 2 x + 3 d x + ∫ d x x 2 + 2 x + 1 2 - 1 2 + 3 I = I = ∫ 1 2 2 x + 2 x 2 + 2 x + 3 d x + ∫ d x x + 1 2 + 2 2 Using: ∫ d x x 2 + a 2 = log x + x 2 + a 2 + c and ∫ f ' x f x d x = 2 f x + c I = x 2 + 2 x + 3 + log | x + 1 + x 2 + 2 x + 3 | + c

Evaluate ∫ x 2 x 6 - 2 x 3 - 3

Let I = ∫ x 2 x 6 - 2 x 3 - 3 Let x 3 = t 3 x 2 d x = d t I = ∫ 1 t 2 - 2 t - 3 d t 3 I = 1 3 ∫ d t t 2 - 2 t + 1 2 - 1 2 - 3 I = 1 3 ∫ d t t - 1 2 - 2 2 I = 1 3 log | t - 1 + t - 1 2 - 2 2 | + c I = 1 3 log | t - 1 + t 2 - 2 t - 3 | + c I = 1 3 log | x 3 - 1 + x 6 - 2 x 3 - 3 | + 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 8: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Integrals, Exercise 8: EXERCISE

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