Explanation Given: The integral: I = ∫ 1 ∞ d x x ( 2 x + 5 ) Formula used: I = ∫ a b f ( x ) d x = [ F ( x ) ] a b = F ( b ) − F ( a ) I = ∫ a b f ( x ) d x = ∫ a c 1 f ( x ) d x + ∫ c 1 c 2 f ( x ) d x + ... + ∫ c n − 1 b f ( x ) d x , Where a < c 1 < c 2 < ... < c n − 1 < b Calculation: The integral we have I = ∫ 1 ∞ d x x ( 2 x + 5 ) Assume, 1 x ( 2 x + 5 ) = A x + B ( 2 x + 5 ) Now, A = lim x → 0 x [ 1 x ( 2 x + 5 ) ] = lim x → 0 [ 1 ( 2 x + 5 ) ] = [ 1 ( 2 × 0 + 5 ) ] = 1 5 B = lim x → − 5 2 ( 2 x + 5 ) [ 1 x ( 2 x + 5 ) ] = lim x → − 5 2 [ 1 x ] = [ 1 − 5 2 ] = − 2 5 ∴ 1 x ( 2 x + 5 ) = A x + B ( 2 x + 5 ) = 1 5 x + − 2 5 ( 2 x + 5 ) (1) I = ∫ 1 ∞ d x x ( 2 x + 5 ) = ∫ 1 ∞ [ 1 5 x + − 2 5 ( 2 x + 5 ) ] d x = ∫ 1 ∞ [ 1 5 x

4th Edition

ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 8.7, Problem 42E

To determine

The value of the integral

$I = \int_{1}^{\infty} \frac{d x}{x (2 x + 5)}$

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Chapter 8.7, Problem 41E

Chapter 8.7, Problem 43E

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Chapter 8 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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