Evaluate the following : ∫ x x + 1 - ln ( x + 1 ) x ( ln ( x + 1 ) ) d x

∫ x x + 1 - ln ( x + 1 ) x ( ln ( x + 1 ) ) d x = ∫ x x + 1 x ( ln ( x + 1 ) ) d x - ∫ ln ( x + 1 ) x ( ln ( x + 1 ) ) d x = ∫ 1 x + 1 ( ln ( x + 1 ) ) d x - ∫ 1 x d x = I 1 - I 2 For I 1 : Let ln x + 1 = t ⇒ d x x + 1 = d t So, I 1 = ∫ d t t = ln t + c 1 = ln ln x + 1 + c 1 and I 2 = ∫ d x x = ln x + c 2 So, I = ln ln x + 1 - ln x + c = ln ln ( x + 1 ) x + c

Integrate with respect to x : 3 x + 2 ( x + 1 ) 2 ( x + 2 )

Let 3 x + 2 ( x + 1 ) 2 ( x + 2 ) ≡ A x + 1 + B x + 1 2 + C x + 2 i.e. 3 x + 2 ( x + 1 ) 2 ( x + 2 ) ≡ A x + 1 x + 2 + B x + 2 + C x + 1 2 x + 1 2 x + 2 i.e. 3 x + 2 ( x + 1 ) 2 ( x + 2 ) ≡ A x 2 + 3 x + 2 + B x + 2 B + C x 2 + 2 x + 1 x + 1 2 x + 2 i.e. 3 x + 2 ≡ x 2 A + C + x 3 A + B + 2 C + 2 A + 2 B + C Comparing the coefficients on both sides, we get A = - C . . . ( 1 ) 3 A + B + 2 C = 3 . . . ( 2 ) 2 A + 2 B + C = 2 . . . ( 3 ) From these equations, A = 4 , B = - 1 , C = - 4 . ∴ ∫ 3 x + 2 ( x + 1 ) 2 ( x + 2 ) d x = ∫ 4 d x x + 1 - ∫ d x x + 1 2 - ∫ 4 d x x + 2 = 4 ln x + 1 + 1 x + 1 - 4 ln x + 2 + c

Evaluate the following : ∫ 1 + x x d x

Let x = tan 2 y ⇒ d x = 2 tan y sec 2 y d y So, I = ∫ 1 + x x d x = 2 ∫ 1 + tan 2 y tan 2 y · tan y · sec 2 y d y = 2 ∫ sec 2 y tan 2 y · tan y · sec 2 y d y = 2 ∫ cosec y · tan y · sec 2 y d y = 2 ∫ sec y · sec 2 y d y Now, Applying integration by parts i.e. ∫ u · v d x = u ∫ v d x - ∫ d u d x ∫ v d x d x , we get I = 2 sec y ∫ sec 2 y d y - ∫ d sec y d y ∫ sec 2 y d y d y ⇒ I = 2 sec y tan y - ∫ sec y tan y · tan y d y ⇒ I = 2 sec y tan y - ∫ sec y tan 2 y d y ⇒ I = 2 sec y tan y - ∫ sec y sec 2 y - 1 d y ⇒ I = 2 sec y tan y - ∫ sec 3 y - sec y d y ⇒ I = 2 sec y tan y - ∫ sec 3 y d y + ∫ sec y d y ⇒ I = 2 sec y tan y - I + 2 ∫ sec y d y ⇒ I = sec y tan y + ∫ sec y d y = sec y tan y + ln sec y + tan y + c ⇒ I = 1 + x · x + ln 1 + x + x + c ⇒ I = x + x 2 + ln 1 + x + x + c

Evaluate: ∫ x 1 + x 1 - x d x

Let I = ∫ x 1 + x 1 - x d x Put x = cos 2 t ⇒ d x = - 2 sin 2 t d t ∴ I = ∫ cos 2 t 1 + cos 2 t 1 - cos 2 t - 2 sin 2 t d t ⇒ I = - 2 ∫ cos 2 t 1 + 2 cos 2 t - 1 1 - 1 + 2 sin 2 t 2 sin t cos t d t ⇒ I = - 4 ∫ cos 2 t cos t sin t sin t cos t d t ⇒ I = - 4 ∫ cos 2 t cos 2 t d t ⇒ I = - 2 ∫ cos 2 t 1 + cos 2 t d t ⇒ I = - 2 ∫ cos 2 t + cos 2 2 t d t ⇒ I = - 2 ∫ cos 2 t + 1 + cos 4 t 2 d t ⇒ I = - ∫ 2 cos 2 t + 1 + cos 4 t d t ⇒ I = - sin 2 t + t + 1 4 sin 4 t + C ⇒ I = - 1 - x 2 + 1 2 cos - 1 x + 1 2 sin 2 t cos 2 t + C ⇒ I = - 1 - x 2 + 1 2 cos - 1 x + x 2 1 - x 2 + C

Evaluate the following: ∫ sin x + cos x 9 + 16 sin 2 x d x

Let sin x - cos x = t ⇒ cos x + sin x d x = d t Also, t 2 = sin x - cos x 2 = sin 2 x + cos 2 x - 2 sin x cos x = 1 - sin 2 x i.e. sin 2 x = 1 - t 2 So, I = ∫ sin x + cos x 9 + 16 sin 2 x d x = ∫ d t 9 + 16 1 - t 2 = - 1 16 ∫ d t t 2 - 5 4 2 Now using ∫ d x x 2 - a 2 = 1 2 a ln x - a x + a + c , we get I = - 1 16 · 1 2 · 5 4 ln t - 5 4 t + 5 4 + c ⇒ I = - 1 40 ln 4 t - 5 4 t + 5 + c ⇒ I = 1 40 ln 4 t + 5 4 t - 5 + c ∵ ln 1 x = - ln x ⇒ I = 1 40 ln 4 sin x - cos x + 5 4 sin x - cos x - 5 + c

If I n = ∫ 1 x 2 + a 2 n d x then prove that I n = x 2 ( n - 1 ) x 2 + a 2 n - 1 + 2 n - 3 2 ( n - 1 ) I n - 1 , n ∈ N , n ≥ 2

Given I n = ∫ 1 x 2 + a 2 n d x , n ∈ N , n ≥ 2 Let x = a tan y ⇒ d x = a sec 2 y d y So, I n = ∫ a sec 2 y a 2 n sec 2 n y d y = 1 a 2 n - 1 ∫ cos 2 n y · sec 2 y d y Applying ∫ u · v d x = u ∫ v d x - ∫ d u d x ∫ v d x d x and taking u = cos 2 n y & v = sec 2 y , ⇒ I n = 1 a 2 n - 1 cos 2 n y ∫ sec 2 y d y - ∫ d cos 2 n y d y ∫ sec 2 y d y d y = 1 a 2 n - 1 cos 2 n y · tan y - ∫ 2 n · cos 2 n - 1 y · - sin y · tan y d y = 1 a 2 n - 1 cos 2 n y · tan y + 2 n ∫ cos 2 n - 2 y · sin 2 y d y = 1 a 2 n - 1 cos 2 n y · tan y + 2 n ∫ cos 2 n - 2 y · 1 - cos 2 y d y = 1 a 2 n - 1 cos 2 n y · tan y + 2 n ∫ cos 2 n - 2 y d y - 2 n ∫ cos 2 n y d y = cos 2 n y · tan y a 2 n - 1 + 2 n · I n - 2 n · I n + 1 = 1 1 + tan 2 y n · x a a 2 n - 1 + 2 n · I n - 2 n · I n + 1 = a 2 a 2 + x 2 n · x a a 2 n - 1 + 2 n · I n - 2 n · I n + 1 ⇒ I n + 1 = x 2 n x 2 + a 2 n + 2 n - 1 2 n I n Replacing n → n - 1 , we get ⇒ I n = x 2 n - 1 x 2 + a 2 n - 1 + 2 n - 3 2 n - 1 I n - 1 , n ∈ N , n ≥ 2

If I n = ∫ x n ( a - x ) 1 2 d x then prove that I n = 2 a n 2 n + 3 I n - 1 - 2 x n ( a - x ) 3 2 2 n + 3

Given I n = ∫ x n ( a - x ) 1 2 d x . Using integration by parts, we get I n = x n ∫ ( a - x ) 1 2 d x - ∫ d x n d x ∫ ( a - x ) 1 2 d x = - 2 x n ( a - x ) 3 2 3 d x + 2 n 3 ∫ x n - 1 ( a - x ) 3 2 d x = - 2 x n ( a - x ) 3 2 3 d x + 2 n 3 ∫ x n - 1 ( a - x ) 1 2 a - x d x = - 2 x n ( a - x ) 3 2 3 d x + 2 a n 3 ∫ x n - 1 ( a - x ) 1 2 d x - 2 n 3 ∫ x n ( a - x ) 1 2 d x = - 2 x n ( a - x ) 3 2 3 d x + 2 a n 3 I n - 1 - 2 n 3 I n ⇒ I n 1 + 2 n 3 = - 2 x n ( a - x ) 3 2 3 d x + 2 a n 3 I n - 1 ⇒ I n = - 2 x n ( a - x ) 3 2 2 n + 3 d x + 2 a n 2 n + 3 I n - 1

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1

Embibe Experts Solutions for Chapter: Indefinite Integration, Exercise 1: Exercise-1

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Indefinite Integration, Exercise 1: Exercise-1

Attempt the free practice questions on Chapter 29: Indefinite Integration, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Indefinite Integration, Exercise 1: Exercise-1 with Hints & Solutions

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 30

Evaluate the following :

$\int (\frac{\frac{x}{x + 1} - \ln (x + 1)}{x (\ln (x + 1))}) d x$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 54

Integrate with respect to $x$ :

$\frac{3 x + 2}{(x + 1)^{2} (x + 2)}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 63

Evaluate the following : $\int \sqrt{(\frac{1 + x}{x})} d x$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 65

Evaluate: $\int (\frac{x \sqrt{1 + x}}{\sqrt{1 - x}}) d x$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 73

Evaluate the following: $\int (\frac{\sin x + \cos x}{9 + 16 \sin 2 x}) d x$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 76

If $I_{n} = \int \frac{1}{{(x^{2} + a^{2})}^{n}} d x$ then prove that $I_{n} = \frac{x}{2 (n - 1) {(x^{2} + a^{2})}^{n - 1}} + \frac{2 n - 3}{2 (n - 1)} I_{n - 1}, n \in N, n \geq 2$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 77

If $I_{n} = \int x^{n} (a - x)^{\frac{1}{2}} d x$ then prove that $I_{n} = \frac{2 a n}{2 n + 3} I_{n - 1} - \frac{2 x^{n} (a - x)^{\frac{3}{2}}}{2 n + 3}$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 29 - Indefinite Integration>Exercise-1>Q 109

If $I_{n} = \int \frac{e^{x}}{x^{n}} d x$ and $I_{n} = \frac{- e^{x}}{k_{1} x^{n - 1}} + \frac{1}{k_{2} - 1} I_{n - 1}, n \in N, n \geq 2$ , then $(k_{2} - k_{1})$ is equal to:

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