Integrate the following w.r.t. x : x 2 + x - 1 x 2 + x - 6

Let I = ∫ x 2 + x - 1 x 2 + x - 6 d x ⇒ I = ∫ 1 + 5 x 2 + x - 6 d x ⇒ I = ∫ 1 + 5 ( x - 2 ) ( x + 3 ) d x We express 5 ( x − 2 ) ( x + 3 ) = A x − 2 + B x + 3 ⇒ 5 = ( x − 2 ) B + ( x + 3 ) A ⇒ 5 = x ( A + B ) + 3 A − 2 B ⇒ A + B = 0 , 3 A − 2 B = 5 Solving above equations, we get A = 1 , B = − 1 ⇒ 5 ( x − 2 ) ( x + 3 ) = 1 x − 2 - 1 x + 3 ∴ I = ∫ 1 + 1 x − 2 - 1 x + 3 d x ⇒ I = ∫ 1 d x + ∫ 1 x − 2 d x - ∫ 1 x + 3 d x ⇒ I = x + ln x - 2 - ln x + 3 + c which is the required solution.

Integrate the following w.r.t. x : ∫ x 2 - 2 x + 3 x 4 d x

I = ∫ x 2 - 2 x + 3 x 4 d x = ∫ x 2 x 4 d x - 2 ∫ x x 4 d x + ∫ 3 x 4 d x = ∫ x - 2 d x - 2 ∫ x - 3 d x + 3 x - 4 + 1 - 4 + 1 + C = - 1 x + 1 x 2 - 1 x 3 + C

Integrate the following w.r.t. x : 2 x 2 + x 2 3 + x 2

Let I = ∫ 2 x 2 + x 2 3 + x 2 d x . . . ( 1 ) Express integral function: 2 x x 2 + 2 x 2 + 3 = A x + B x 2 + 2 + C x + D x 2 + 3 ⇒ 2 x x 2 + 2 x 2 + 3 = x 2 + 2 ( C x + D ) + x 2 + 3 ( A x + B ) x 2 + 2 x 2 + 3 ⇒ 2 x = x 2 + 2 ( C x + D ) + x 2 + 3 ( A x + B ) ⇒ 2 x = x 3 A + x 3 C + x 2 B + x 2 D + 3 x A + 2 x C + 3 B + 2 D ⇒ 2 x = x 3 ( A + C ) + x 2 ( B + D ) + x ( 3 A + 2 C ) + 3 B + 2 D ⇒ A + C = 0 , B + D = 0 , 3 A + 2 C = 2 and 3 B + 2 D = 0 Solving above equations simultaneously, we get A = 2 , B = 0 , C = - 2 and D = 0 ⇒ 2 x x 2 + 2 x 2 + 3 = 2 x x 2 + 2 + − 2 x x 2 + 3 Equation ( 1 ) becomes, I = ∫ 2 x x 2 + 2 + − 2 x x 2 + 3 d x ⇒ I = ∫ 2 x x 2 + 2 d x - ∫ 2 x x 2 + 3 d x ⇒ I = ln x 2 + 2 - ln x 2 + 3 + c which is the required solution.

Integrate the following w.r.t. x : 2 x + 3 x 2 + 4 x + 1

Let, I = ∫ 2 x + 3 x 2 + 4 x + 1 d x Let, 2 x + 3 = A d x 2 + 4 x + 1 d x + B ⇒ 2 x + 3 = A ( 2 x + 4 ) + B ⇒ 2 x + 3 = ( 2 A ) x + 4 A + B Comparing coefficient of x and constant, we get ⇒ 2 = 2 A ⇒ A = 1 And, 3 = 4 A + B ⇒ 3 = 4 + B ⇒ B = - 1 Now I = ∫ [ ( 2 x + 4 ) - 1 ] x 2 + 4 x + 1 · d x I = ∫ ( 2 x + 4 ) · d x x 2 + 4 x + 1 - ∫ d x x 2 + 4 x + 1 I = I 1 - I 2 . . . . . . . . . . . . . . . . . . . . . . . i Now, I 1 = ∫ ( 2 x + 4 ) · d x x 2 + 4 x + 1 Let, x 2 + 4 x + 1 = t ⇒ ( 2 x + 4 ) d x = 4 t ⇒ I 1 = ∫ d t t ⇒ I 1 = t 1 2 ⇒ I 1 = 2 x 2 + 4 x + 1 And, I 2 = ∫ d x x 2 + 4 x + 1 I 2 = ∫ d x ( x + 2 ) 2 + 1 - 4 I 2 = ∫ d x ( x + 2 ) 2 - 3 I 2 = ∫ d x ( x + 2 ) 2 - ( 3 ) 2 By using the formula, ∫ d x x 2 - a 2 = log x + x 2 - a 2 + c I 2 = log ( x + 2 ) + ( x + 2 ) 2 - 3 I 2 = log ( x + 2 ) + x 2 + 4 x + 1 From equation i , I = 2 x 2 + 4 x + 1 - log ( x + 2 ) + x 2 + 4 x + 1 + c Hence, ∫ 2 x + 3 x 2 + 4 x + 1 d x = 2 x 2 + 4 x + 1 - log ( x + 2 ) + x 2 + 4 x + 1 + c

Integrate the following w.r.t. x : ∫ x 2 + 3 x - 1 x + 1 2 d x

Let I = ∫ x 2 + 3 x - 1 ( x + 1 ) 2 d x = ∫ x 2 + x + 2 x - 1 ( x + 1 ) 2 d x = ∫ x ( x + 1 ) + 2 x - 1 ( x + 1 ) 2 d x = ∫ x ( x + 1 ) ( x + 1 ) 2 d x + ∫ 2 x - 1 ( x + 1 ) 2 d x = ∫ x x + 1 d x + ∫ 2 x + 2 - 2 - 1 x + 1 2 d x = ∫ x + 1 - 1 x + 1 d x + ∫ 2 ( x + 1 ) - 3 ( x + 1 ) 2 d x = ∫ x + 1 x + 1 d x - ∫ 1 x + 1 d x + ∫ 2 ( x + 1 ) ( x + 1 ) 2 d x - 3 ∫ 1 ( x + 1 ) 2 d x = ∫ d x - ∫ 1 x + 1 d x + 2 ∫ 1 ( x + 1 ) d x - 3 ∫ 1 ( x + 1 ) 2 d x = ∫ d x + ∫ 1 x + 1 d x - 3 ∫ 1 ( x + 1 ) 2 d x = x + log x + 1 + 3 x + 1 + C

Integrate the following w.r.t. x : 5 x 2 + 20 x + 6 x 3 + 2 x 2 + x

Let I = ∫ 5 x 2 + 20 x + 6 x 3 + 2 x 2 + x d x We express 5 x 2 + 20 x + 6 x ( x + 1 ) 2 = A x + B x + 1 + C ( x + 1 ) 2 ⇒ 5 x 2 + 20 x + 6 = x ( x + 1 ) B + x C + ( x + 1 ) 2 A ⇒ 5 x 2 + 20 x + 6 = x 2 ( A + B ) + x ( 2 A + B + C ) + A ⇒ A + B = 5 , 2 A + B + C = 20 , A = 6 Solving above equations, we get A = 6 , B = − 1 , C = 9 ⇒ 5 x 2 + 20 x + 6 x 3 + 2 x 2 + x = 6 x - 1 x + 1 + 9 ( x + 1 ) 2 ∴ I = ∫ 6 x - 1 x + 1 + 9 ( x + 1 ) 2 d x = ∫ 6 x d x - ∫ 1 x + 1 d x + ∫ 9 ( x + 1 ) 2 d x = 6 ln | x | - ln | x + 1 | - 9 x + 1 + c which is the required solution.

Integrate the following w.r.t. x : 1 x 2 + 4 x - 5

Given, ∫ 1 x 2 + 4 x - 5 d x 1 x 2 + 4 x - 5 = 1 ( x - 1 ) ( x + 5 ) Applying partial decomposition, Let 1 ( x - 1 ) ( x + 5 ) = A x - 1 + B x + 5 ⇒ 1 ( x - 1 ) ( x + 5 ) = ( x - 1 ) B + ( x + 5 ) A ( x - 1 ) ( x + 5 ) ⇒ 1 = ( x - 1 ) B + ( x + 5 ) A ⇒ 1 = x ( A + B ) + 5 A - B Comparing corresponding coefficients, we get A + B = 0 & 5 A - B = 1 Solving above equations, we get A = 1 6 , B = - 1 6 ⇒ ∫ 1 x 2 + 4 x - 5 d x = ∫ 1 6 ( x - 1 ) - 1 6 ( x + 5 ) d x = ∫ 1 6 ( x - 1 ) d x - ∫ 1 6 ( x + 5 ) d x = 1 6 ∫ 1 ( x - 1 ) d x - 1 6 ∫ 1 ( x + 5 ) d x If, ∫ f x d x = F x + C , then ∫ f a x + b d x = 1 a F a x + b + C . ⇒ 1 6 ∫ 1 ( x - 1 ) d x - 1 6 ∫ 1 ( x + 5 ) d x = 1 6 ln | x - 1 | - 1 6 ln | x + 5 | ∴ 1 x 2 + 4 x - 5 = 1 6 ln | x - 1 | - 1 6 ln | x + 5 | + c

Integrate the following w.r.t. x : 2 log x + 3 x 3 log x + 2 log x 2 + 1

Let l = ∫ 2 log ⁡ x + 3 x ( 3 log ⁡ x + 2 ) [ ( log ⁡ x ) 2 + 1 ] ⋅ d x P u t log ⁡ x = t ∴ 1 x ⋅ d x = dt ∴ I = ∫ 2 t + 3 ( 3 t + 2 ) ( t 2 + 1 ) ⋅ d t Let 2 t + 3 ( 3 t + 2 ) ( t 2 + 1 ) = A 3 t + 2 + Bt + C t 2 + 1 ∴ 2 t + 3 = A ( t 2 + 1 ) + ( Bt + C ) ( 3 t + 2 ) P u t 3 t + 2 = 0 i , e , t = − 2 3 , w e g e t 2 ( − 2 3 ) + 3 = A ( 4 9 + 1 ) + ( − 2 3 B + C ) ( 0 ) ∴ 5 3 = A ( 13 9 ) ∴ A = 15 13 P u t t = 0 3 = A ( 1 ) + C ( 2 ) = 15 13 + 2 C ∴ 2 C = 3 − 15 13 = 24 13 ∴ C = 12 13 C o m p a r i n g c o e f f i c i e n t o f t 2 on both the sides , we get 0 = A + 3 B ∴ B = − A 3 = − 5 13 ∴ 2 t + 3 ( 3 t + 2 ) ( t 2 + 1 ) = ( 15 13 ) 3 t + 2 + ( − 5 13 t + 2 13 ) t 2 + 1 ∴ I = ∫ [ ( 15 13 ) 3 t + 2 + ( − 5 13 t + 12 3 ) t 2 + 1 ] ⋅ dt = 15 13 ∫ 1 3 t + 2 ⋅ dt − 5 26 ∫ 2 t t ⋅ 2 + 1 ⋅ dt + 12 13 ∫ 1 t 2 + 1 ⋅ dt = 15 13 ⋅ 1 3 log ⁡ | 3 t + 2 | − 5 26 log ⁡ | t 2 + 1 | + 12 13 tan − 1 ⁡ ( t ) + c ⋯ [ ∵ d dt ( t 2 + 1 ) = 2 t and ∫ f ′ ( x ) f ( x ) dt = log ⁡ | f ( t ) | + c ] = 5 13 log ⁡ | 3 log ⁡ x + 2 | − 5 26 log ⁡ | ( log ⁡ x ) 2 + 1 | + 12 13 tan − 1 ⁡ ( log ⁡ x ) + c .

E M B I B E

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 3 - Indefinite Integration>EXERCISE 3.4

Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 6: EXERCISE 3.4

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Indefinite Integration, Exercise 6: EXERCISE 3.4

Attempt the practice questions on Chapter 3: Indefinite Integration, Exercise 6: EXERCISE 3.4 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.