Evaluate the following integral: ∫ 3 - 2 x - 2 x 2 d x .

Let I = ∫ 3 - 2 x - 2 x 2 d x ⇒ I = ∫ 2 3 2 - x - x 2 d x ⇒ I = 2 ∫ 7 4 - 1 4 - x - x 2 d x ⇒ I = 2 ∫ 7 2 2 - x + 1 2 2 d x Using the standard formula: ∫ a 2 - x 2 d x = x 2 a 2 - x 2 + a 2 2 sin - 1 x a + C ∴ I = 2 x + 1 2 2 3 2 - x - x 2 + 7 4 × 2 sin - 1 x + 1 2 7 2 + C Hence, ∫ 3 - 2 x - 2 x 2 d x = 1 4 ( 2 x + 1 ) 3 - 2 x - 2 x 2 + 7 4 2 sin - 1 2 x + 1 7 + C .

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

Let I = ∫ x x 4 + 1 d x ⇒ I = ∫ x x 2 2 + 1 d x Substitute x 2 = t Differentiate with respect to x , then we get: 2 x d x = d t ∴ I = 1 2 ∫ t 2 + 1 d t ⇒ I = 1 2 ∫ t 2 + 1 2 d t ⇒ I = 1 2 t 2 t 2 + 1 + 1 2 2 ln | t + t 2 + 1 | + C ⇒ I = 1 2 x 2 2 x 4 + 1 + 1 2 ln x 2 + x 4 + 1 + C , where t = x 2 Hence, ∫ x x 4 + 1 d x = x 2 4 x 4 + 1 + 1 4 ln x 2 + x 4 + 1 + C .

Evaluate the following integral: ∫ x 2 a 6 - x 6 d x .

Let I = ∫ x 2 a 6 - x 6 d x ⇒ I = ∫ a 3 2 - x 3 2 x 2 d x Substitute x 3 = t Differentiate with respect to x , then we get: 3 x 2 d x = d t Therefore, I = 1 3 ∫ a 3 2 - t 2 d t Using the standard formula: ∫ a 2 - x 2 d x = x 2 a 2 - x 2 + a 2 2 sin - 1 x a + C Thus, I = 1 3 t 2 a 3 2 - t 2 + a 3 2 2 sin - 1 t a 3 + C Hence, ∫ x 2 a 6 - x 6 d x = 1 6 x 3 a 6 - x 6 + a 6 6 sin - 1 x 3 a 3 + C , where t = x 3

Evaluate the following integral: ∫ 16 + ( log x ) 2 x d x .

Let I = ∫ 16 + ( log x ) 2 x d x Substitute log x = t Differentiate with respect to x , then we get: 1 x d x = d t ∴ I = ∫ 16 + t 2 d t ⇒ I = ∫ ( 4 ) 2 + t 2 d t ⇒ I = t 2 4 2 + t 2 + 16 2 ln t + 4 2 + t 2 + C [Using ∫ a 2 + x 2 d x = x 2 a 2 + x 2 + a 2 2 ln x + a 2 + x 2 + C ] Hence, ∫ 16 + ( log x ) 2 x d x = 1 2 log x ( log x ) 2 + 16 + 8 ln log x + ( log x ) 2 + 16 + C , where t = log x

Evaluate the following integral: ∫ 2 a x - x 2 d x .

Let I = ∫ 2 a x - x 2 d x ⇒ I = ∫ a 2 - x 2 - 2 a x + a 2 d x ⇒ I = ∫ a 2 - ( x - a ) 2 d x Using the standard formula: ∫ a 2 - x 2 d x = x 2 a 2 - x 2 + a 2 2 sin - 1 x a + C Therefore, ∫ 2 a x - x 2 d x = 1 2 ( x - a ) a 2 - ( x - a ) 2 + a 2 2 sin - 1 x - a a + C = 1 2 ( x - a ) 2 a x - x 2 + a 2 2 sin - 1 x - a a + C .

Evaluate the following integral: ∫ 3 - x 2 d x .

Let I = 3 - x 2 d x ⇒ I = ∫ ( 3 ) 2 - x 2 d x Using the standard formula: ∫ a 2 - x 2 d x = x 2 a 2 - x 2 + a 2 2 sin - 1 x a + C Therefore, I = x 2 ( 3 ) 2 - x 2 + 3 2 2 sin - 1 x 3 + C Hence, ∫ 3 - x 2 d x = 1 2 x 3 - x 2 + 3 2 sin - 1 x 3 + C .

Evaluate the following integral: ∫ x 2 - 2 x d x .

Let I = ∫ x 2 - 2 x d x ⇒ I = ∫ x 2 - 2 x + 1 - 1 d x ⇒ I = ∫ ( x - 1 ) 2 - ( 1 ) 2 d x Using the standard formula: ∫ x 2 - a 2 d x = x 2 x 2 - a 2 - a 2 2 ln x + x 2 - a 2 + C Therefore, I = x - 1 2 ( x - 1 ) 2 - ( 1 ) 2 - 1 2 ln x - 1 + ( x - 1 ) 2 - ( 1 ) 2 + C Hence, ∫ x 2 - 2 x d x = 1 2 x - 1 x 2 - 2 x - 1 2 ln x - 1 + x 2 - 2 x + C .

Evaluate the following integral: ∫ 2 x - x 2 d x .

Let I = ∫ 2 x - x 2 d x ⇒ I = ∫ 1 - 1 + 2 x - x 2 d x ⇒ I = ∫ ( 1 ) 2 - x 2 - 2 x + 1 d x ⇒ I = ∫ ( 1 ) 2 - ( x - 1 ) 2 d x Using the standard formula: ∫ a 2 - x 2 d x = x 2 a 2 - x 2 + a 2 2 sin - 1 x a + C Therefore, I = x - 1 2 ( 1 ) 2 - ( x - 1 ) 2 + 1 2 sin - 1 x - 1 1 + C Hence, ∫ 2 x - x 2 d x = 1 2 x - 1 2 x - x 2 + 1 2 sin - 1 ( x - 1 ) + C .

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 19 - Indefinite Integrals>EXERCISE

R D Sharma Solutions for Chapter: Indefinite Integrals, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Indefinite Integrals, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 19: Indefinite Integrals, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.