∫ d x ( 1 + x ) 3 x 2 + 3 x + 1

Let I = ∫ d x 1 + x 3 x 2 + 3 x + 1 Let x + 1 = 1 t ⇒ d x = - 1 t 2 d t ∴ I = ∫ - 1 t 2 d t 1 t 3 1 t - 1 2 + 3 1 t - 1 + 1 = ∫ - d t t 2 - 3 t + 3 = ∫ - d t t - 3 2 2 + 3 4 = - log t - 3 2 + t 2 - 3 t + 3 + C = - log 1 x + 1 - 3 2 + 1 x + 1 2 - 3 x + 1 + 3 + C = - log - 3 x - 1 2 ( x + 1 ) + 3 x 2 + 3 x + 1 x + 1 + C = - log - 3 x - 1 + 2 3 x 2 + 3 x + 1 2 ( x + 1 ) + C , where C is constant of integration.

∫ d x ( 1 + x ) 3 x 2 + 3 x + 1

Let I = ∫ d x 1 + x 3 x 2 + 3 x + 1 Let x + 1 = 1 t ⇒ d x = - 1 t 2 d t ∴ I = ∫ - 1 t 2 d t 1 t 3 1 t - 1 2 + 3 1 t - 1 + 1 = ∫ - d t t 2 - 3 t + 3 = ∫ - d t t - 3 2 2 + 3 4 = - log t - 3 2 + t 2 - 3 t + 3 + C = - log 1 x + 1 - 3 2 + 1 x + 1 2 - 3 x + 1 + 3 + C = - log - 3 x - 1 2 ( x + 1 ) + 3 x 2 + 3 x + 1 x + 1 + C = - log - 3 x - 1 + 2 3 x 2 + 3 x + 1 2 ( x + 1 ) + C , where C is constant of integration.

∫ d x ( 3 + x 2 ) 1 + x

Let I = ∫ d x ( 3 + x 2 ) 1 + x Let 1 + x = t 2 ⇒ d x = 2 t d t ∴ I = ∫ 2 t d t 3 + ( t 2 - 1 ) 2 t 2 = ∫ 2 d t t 4 - 2 t 2 + 4 = 1 2 ∫ 4 d t t 4 - 2 t 2 + 4 = 1 2 ∫ ( 2 + t 2 ) + ( 2 - t 2 ) ( t 4 - 2 t 2 + 4 ) d t = 1 2 ∫ 2 + t 2 t 4 - 2 t 2 + 4 + 2 - t 2 t 4 - 2 t 2 + 4 d t Divide N r & D r by t 2 , = 1 2 ∫ 2 t 2 + 1 t 2 - 2 + 4 t 2 + 2 t 2 - 1 t 4 - 2 + 4 t 2 d t = 1 2 ∫ 1 + 2 t 2 t - 2 t 2 + 2 - ( 1 - 2 t 2 ) ( t + 2 t ) 2 - 6 d t = 1 2 I 1 - 1 2 I 2 + C For I 1 = ∫ 1 + 2 t 2 t - 2 t 2 + 2 2 d t Let t - 2 t = z ⇒ 1 + 2 t 2 d t = d z ∴ I 1 = ∫ d z z 2 + ( 2 ) 2 = 1 2 tan - 1 z 2 = 1 2 tan - 1 t - 2 t 2 And for I 2 = ∫ 1 - 2 t 2 t + 2 t 2 - ( 6 ) 2 d t Let t + 2 t = u ⇒ 1 - 2 t 2 d t = d u ∴ I 2 = ∫ d z u 2 - ( 6 ) 2 = 1 2 6 log u - 6 u + 6 = 1 2 6 log t + 2 t - 6 t + 2 t + 6 ∴ I = 1 2 I 1 - 1 2 I 2 + C = 1 2 2 tan - 1 t 2 - 2 2 t - 1 4 6 log t 2 + 2 - 6 t t 2 + 2 + 6 t + C = 1 2 2 tan - 1 x + 1 - 2 2 x + 1 - 1 4 6 log x + 3 - 6 x + 1 x + 3 + 6 x + 1 + C = 1 2 2 tan - 1 x - 1 2 x + 2 + 1 4 6 log x + 3 + 6 x + 6 x + 3 - 6 x + 6 + C

Let I = ∫ d x x 2 x 2 - 1 Let x = 1 t ⇒ d x = - 1 t 2 d t ∴ I = ∫ - 1 t 2 d t 1 t 2 1 t 2 - 1 = = ∫ - t d t 1 - t 2 Again let 1 - t 2 = z 2 ⇒ - 2 t d t = 2 z d t ⇒ - t d t = z d z ∴ I = ∫ z d z z 2 = z + C = 1 - t 2 + C = 1 - 1 x 2 + C = x 2 - 1 x + C , where C is constant of integration.

∫ d x ( x 2 - 1 ) x 2 + 1

Let I = ∫ d x ( x 2 - 1 ) x 2 + 1 Let x = 1 t ⇒ d x = - 1 t 2 d t ∴ I = ∫ - 1 t 2 d t 1 t 2 - 1 1 t 2 + 1 = ∫ - t d t ( 1 - t 2 ) 1 + t 2 Let 1 + t 2 = z 2 ⇒ 2 t d t = 2 z d z ⇒ t d t = z d z ∴ I = ∫ - z d z 1 - ( z 2 - 1 ) z 2 = ∫ d z z 2 - 2 = 1 2 2 log z - 2 z + 2 + C = 1 2 2 log 1 + t 2 - 2 1 + t 2 + 2 + C = 1 2 2 log x 2 + 1 - x 2 x 2 + 1 + 2 x + C , where C is constant of integration.

∫ x 3 1 + x 2 1 3 d x

Let I = ∫ x 3 1 + x 2 1 3 d x = ∫ x 2 · 1 + x 2 1 3 · x d x Let 1 + x 2 = t ⇒ 2 x d x = d t ∴ I = ∫ t - 1 · t 1 3 · d t 2 = 1 2 ∫ t 4 3 - t 1 3 d t = 1 2 t 4 3 + 1 4 3 + 1 - t 1 3 + 1 1 3 + 1 + C = 3 14 1 + x 2 7 3 - 3 8 1 + x 2 4 3 + C = 3 56 1 + x 2 4 3 ( 4 1 + x 2 - 7 ) + C = 3 56 1 + x 2 4 3 4 x 2 - 3 + C , where C is constant of integration.

If I n = ∫ d x x 2 + a 2 n , n ∈ N , then show that: I n + 1 = 1 2 n a 2 x x 2 + a 2 + 2 n - 1 2 n · 1 a 2 I n .

Given I n = ∫ d x x 2 + a 2 n = ∫ x 2 + a 2 - n I · 1 II d x Integrating by parts taking 1 as second function, = x 2 + a 2 - n · x - ∫ - n x 2 + a 2 - n - 1 · 2 x · x d x = x x 2 + a 2 n + 2 n ∫ x 2 x 2 + a 2 n + 1 d x = x x 2 + a 2 n + 2 n ∫ x 2 + a 2 - a 2 x 2 + a 2 n + 1 d x = x x 2 + a 2 n + 2 n ∫ d x x 2 + a 2 - 2 n a 2 ∫ d x x 2 + a 2 n + 1 I n = x x 2 + a 2 n + 2 n I n - 2 n a 2 I n + 1 I n + 1 2 n a 2 = x x 2 + a 2 n + ( 2 n - 1 ) I n ⇒ I n + 1 = 1 2 n a 2 · x x 2 + a 2 n + 2 n - 1 2 n a 2 I n

If I n = ∫ a 2 + x 2 n d x , then Prove that: I n = x a 2 + x 2 n n + 1 + n a 2 n + 1 ∫ a 2 + x 2 n 2 - 1 d x .

I n = ∫ a 2 + x 2 n 2 · 1 II d x Integrating by part taking 1 as II nd function, = a 2 + x 2 n 2 · x - ∫ n 2 a 2 + x 2 n 2 - 1 · 2 x · x d x = a 2 + x 2 n 2 · x - n ∫ x 2 + a 2 n 2 - 1 x 2 + a 2 - a 2 d x = x a 2 + x 2 n 2 - n ∫ x 2 + a 2 n 2 - a 2 x 2 + a 2 n 2 - 1 d x = x a 2 + x 2 n 2 - n I n + n a 2 ∫ x 2 + a 2 n 2 - 1 d x ⇒ 1 + n I n = x a 2 + x 2 n 2 + n a 2 ∫ x 2 + a 2 n 2 - 1 d x ⇒ I n = x ( a 2 + x 2 ) n 2 n + 1 + n a 2 n + 1 ∫ x 2 + a 2 n 2 - 1 d x

If I m , n = ∫ sin m x cos n x d x , then prove that: I m , n = - sin m - 1 x cos n + 1 x m + n + m - 1 m + n I m - 2 , n .

Let y = sin m - 1 x cos n + 1 x ⇒ d y d x = m - 1 sin m - 2 x cos x · cos n + 1 x + n + 1 cos n x · - sin x sin m - 1 x = m - 1 sin m - 2 x cos n x · ( 1 - sin 2 x ) - n + 1 · sin m x cos n x = ( m - 1 ) sin m - 2 x cos n x - ( m - 1 + n + 1 ) sin m x cos n x ⇒ d y d x = m - 1 sin m - 2 x cos n x - m + n sin m x cos n x Integrating w.r.t x on both sides we get y = m - 1 ∫ sin m - 2 x · cos n x d x - m + n ∫ sin m x · cos n x d x ⇒ sin m - 1 x · cos n + 1 x = m - 1 I m - 2 , n - m + n I m , n ⇒ m + n I m , n = - sin m - 1 x · cos n + 1 x + m - 1 I m - 2 , n ⇒ I m , n = - sin m - 1 x · cos n + 1 x m + n + m - 1 m + n I m - 2 , n

If I n = ∫ x cos e c n x d x , n ≥ 3 , then prove that I n = - 1 n - 1 · cos n - 2 x x cot x + 1 + n - 2 n - 1 I n - 2 .

I n = ∫ x cosec n x d x = ∫ x cosec n - 2 x I · cosec 2 x II d x = x cosec n - 2 x ∫ cosec 2 x d x - ∫ d d x x cosec n - 2 x · ∫ cosec 2 x d x d x = x cosec n - 2 x · - cot x - ∫ x · n - 2 cosec n - 3 x · - cosec x cot x + cosec n - 2 x - cot x d x I n = - x cot x cosec n - 2 x - ∫ n - 2 x cosec n - 2 x cot 2 x d x + ∫ cosec n - 2 x cot x d x = - x cot x cosec n - 2 x - n - 2 ∫ x cosec n - 2 x cosec 2 x - 1 d x + ∫ cosec n - 3 x cosec x cot x d x = - x cot x cosec n - 2 x - n - 2 I n + n - 2 I n - 2 + ∫ t n - 3 - d t t = cosec x 1 + n - 2 I n = - x cot x cosec n - 2 x + n - 2 I n - 2 - t n - 3 + 1 n - 3 + 1 n - 1 I n = - x cot x cosec n - 2 x + n - 2 I n - 2 - cosec n - 2 x n - 2 = - cosec n - 2 x x cot x + 1 n - 2 + n - 2 I n - 2 ⇒ I n = - cosec n - 2 x n - 1 x cot x + 1 n - 2 + n - 2 n - 1 I n - 2

E M B I B E

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 94

HARD

JEE Main

IMPORTANT

Earn 100

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

Important Questions on Indefinite Integrals - II

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 94

\int \frac{d x}{(3 + x^{2}) \sqrt{1 + x}}

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 94

\int \frac{d x}{x^{2} \sqrt{x^{2} - 1}}

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 94

\int \frac{d x}{(x^{2} - 1) \sqrt{x^{2} + 1}}

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 94

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

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 95

If

I_{n} = \int \frac{d x}{{(x^{2} + a^{2})}^{n}}, n \in N

, then show that:

I_{n + 1} = \frac{1}{2 n a^{2}} \frac{x}{(x^{2} + a^{2})} + \frac{2 n - 1}{2 n} \cdot \frac{1}{a^{2}} I_{n}

.

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 96

If $I_{n} = \int \sqrt{{(a^{2} + x^{2})}^{n}} d x$ , then

Prove that: $I_{n} = \frac{x \sqrt{{(a^{2} + x^{2})}^{n}}}{n + 1} + \frac{n a^{2}}{n + 1} \int {(a^{2} + x^{2})}^{\frac{n}{2} - 1} d x$ .

HARD

JEE Main

IMPORTANT

Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 7 - Indefinite Integrals - II>EXERCISE>Q 97

If $I_{m, n} = \int \sin^{m} x \cos^{n} x d x$ , then prove that: $I_{m, n} = - \frac{\sin^{m - 1} x \cos^{n + 1} x}{m + n} + \frac{m - 1}{m + n} I_{m - 2, n}$ .