Integrate: ∫ e 2 x - 2 e x + 5 e x d x .

Given, ∫ e 2 x - 2 e x + 5 e x d x = ∫ e x + 5 e - x - 2 d x = ∫ e x d x + 5 ∫ e - x d x - 2 ∫ 1 d x = e x - 5 e - x - 2 x + C ∴ ∫ e 2 x - 2 e x + 5 e x d x = e x - 5 e - x - 2 x + C

Integrate: ∫ 4 e 5 x - 9 e 4 x - 3 e 2 x d x .

Given, ∫ 4 e 5 x - 9 e 4 x - 3 e 2 x d x = ∫ 4 e 3 x - 9 e 2 x - 3 e - 2 x d x = 4 ∫ e 3 x d x - 9 ∫ e 2 x d x - 3 ∫ e - 2 x d x = 4 e 3 x 3 - 9 e 2 x 2 + 3 e - 2 x 2 + C ∴ ∫ 4 e 5 x - 9 e 4 x - 3 e 2 x d x = 4 e 3 x 3 - 9 e 2 x 2 + 3 e - 2 x 2 + C

Integrate: ∫ 5 2 log 5 ( 2 x + 3 ) d x .

Given, ∫ 5 2 log 5 ( 2 x + 3 ) d x = ∫ 5 log 5 ( 2 x + 3 ) 2 d x = ∫ 2 x + 3 2 d x ∵ a log a b = b log a a = b = 2 x + 3 3 3 + C ∵ ∫ a x + b n d x = a x + b n + 1 a n + 1 + C , n ≠ - 1 ∴ ∫ 5 2 log 5 ( 2 x + 3 ) d x = ( 2 x + 3 ) 3 3 + C

Integrate: ∫ e 7 log x - e 5 log x e 3 log x - e log x d x .

Given, ∫ e 7 log x - e 5 log x e 3 log x - e log x d x = ∫ e log x 7 - e log x 5 e log x 3 - e log x d x = ∫ x 7 - x 5 x 3 - x d x ∵ e log a = a = ∫ x 4 x 3 - x x 3 - x d x = ∫ x 4 d x = x 5 5 + C ∴ ∫ e 7 log x - e 5 log x e 3 log x - e log x d x = x 5 5 + C

Evaluate: ∫ 27 1 + x - 9 1 - x 3 x d x .

Given, ∫ 27 1 + x - 9 1 - x 3 x d x = ∫ 3 3 1 + x - 3 2 ( 1 - x ) 3 x d x = ∫ 3 3 + 3 x - 3 2 - 2 x 3 x d x = ∫ 3 3 + 2 x - 3 2 - 3 x d x = ∫ 3 3 · 3 2 x - 3 2 · 3 - 3 x d x = 27 ∫ 3 2 x d x - 9 ∫ 3 - 3 x d x = 27 3 2 x 2 log 3 - 9 3 - 3 x ( - 3 log 3 ) + c ∵ ∫ a b x d x = a b x b log a + c = 27 3 2 x 2 log 3 + 3 3 - 3 x log 3 + c ∴ ∫ 27 1 + x - 9 1 - x 3 x d x = 27 3 2 x 2 log 3 + 3 3 - 3 x log 3 + c

Evaluate: ∫ 8 2 + x + 2 2 x + 3 d x .

Given, ∫ 8 2 + x + 2 2 x + 3 d x = ∫ 8 2 · 8 x + 2 2 x · 2 3 d x = ∫ 64 · 2 3 x + 2 2 x · 8 d x = ∫ 8 ( 2 2 x ) + 1 4 2 - x d x = 8 ∫ 2 2 x d x + 1 4 ∫ 2 - x d x = 8 2 2 x 2 log 2 + 1 4 2 - x - log 2 + c = 4 2 x log 2 - 1 4 2 - x log 2 + c ∴ ∫ 8 2 + x + 2 2 x + 3 d x = 4 2 x log 2 - 1 4 2 - x log 2 + c

Integrate: ∫ a x + b x 2 a x b x d x : a , b > 0 .

Given, ∫ b x + a x 2 a x b x d x = ∫ b 2 x + a 2 x + 2 a x b x a x b x d x = ∫ b x a x + a x b x + 2 d x = ∫ b a x d x + ∫ a b x d x + 2 ∫ 1 d x = b a x 1 log b a + a b x 1 log a b + 2 x + C = b a x 1 log b - log a + a b x 1 log a - log b + 2 x + C ∴ ∫ b x + a x 2 a x b x d x = b a x 1 log b - log a + a b x 1 log a - log b + 2 x + C

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RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - B. Integral Calculus>Chapter 11 - Indefinite Integral>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.