Explanation Given: The integral function is ∫ 3 − x 3 x 2 − 2 x − 1 d x . Formula used: ∫ 1 x d x = ln x + C . Calculation: Consider following indefinite integral, ∫ 3 − x 3 x 2 − 2 x − 1 d x The above integral function can be written as, 3 − x 3 x 2 − 2 x − 1 = 3 − x ( 3 x + 1 ) ( x − 1 ) Now by using partial fraction method we get, 3 − x ( 3 x + 1 ) ( x − 1 ) = A 3 x + 1 + B x − 1 By simplifying further we get, 3 − x = A ( x − 1 ) + B ( 3 x + 1 ) At x = 1 , 3 − 1 = A ( 1 − 1 ) + B ( 3 ( 1 ) + 1 ) B = 1 2 At x = − 1 3 , 3 + 1 3 = A ( − 1 3 − 1 ) + B ( 3 ( − 1 3 ) + 1 )

Student Solutions Manual For Larson/edwards' Calculus Of A Single Variable: Early Transcendental Functions, 2nd

7th Edition

ISBN: 9781337552561

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 8.5, Problem 6E

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To calculate: The indefinite integral for the given function, by using partial fractions.

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Chapter 8.5, Problem 5E

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Chapter 8 Solutions

Student Solutions Manual For Larson/edwards' Calculus Of A Single Variable: Early Transcendental Functions, 2nd

Ch. 8.1 - Integration Technique Describe how to integrate a...Ch. 8.1 - Fitting Integrands to Basic Integration Rules What...Ch. 8.1 - Choosing an Antiderivative In Exercises 3 and 4,...Ch. 8.1 - Choosing an Antiderivative In Exercises 3 and 4,...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...Ch. 8.1 - Choosing a Formula In Exercises 514, select the...

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