Explanation Given information: The integral function is ∫ 1 x 3 − 1 d x . Calculation: Show the integral function as follows: ∫ 1 x 3 − 1 d x (1) Consider the function y = 1 x 3 − 1 (2) Expand Equation (2) as shown below. 1 x 3 − 1 = 1 ( x − 1 ) ( x 2 + x + 1 ) Apply partial fraction decomposition. 1 ( x − 1 ) ( x 2 + x + 1 ) = A x − 1 + B x + C x 2 + x + 1 1 = A ( x 2 + x + 1 ) + ( B x + C ) ( x − 1 ) 1 = A x 2 + A x + A + B x 2 − B x + C x − C 1 = ( A + B ) x 2 + ( A − B + C ) x + ( A − C ) (3) Equating the coefficients to get the value of A , B , and C . A + B = 0 A = − B (4) A − B + C = 0 2 A + C = 0 (5) A − C = 1 (6) Solving Equations (5) and (6). A = 1 3 B = − 1 3 C = − 2 3 Modify Equation (1) using partial fraction decomposition. ∫ 1 x 3 + 1 d x = ∫ ( 1 3 x − 1 + ( − 1 3 ) x + ( − 2 3 ) x 2 + x + 1 ) d x = 1 3 ∫ 1 x − 1 d x − 1 3 ∫ ( x + 1 2 ) x 2 + x + 1 d x 1 − 1 3 ∫ ( 3 2 ) ( x + 1 2 ) 2 + 3 4 d x (7) Consider u = x 2 + x + 1 (8) Differentiate both sides of the Equation (8)

Bundle: Stewart, Essential Calculus: Early Transcendentals, 2nd (hardound) + WebAssign Printed Access Card for Stewart's Essential Calculus: Early ... Multi-Term + WebAssign - Start Smart Guide

2nd Edition

ISBN: 9781133425946

Author: James Stewart

Publisher: Cengage Learning

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Chapter 6.3, Problem 29E

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To evaluate: The integral function ∫1x3−1dx.

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Chapter 6.3, Problem 31E

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Chapter 6 Solutions

Bundle: Stewart, Essential Calculus: Early Transcendentals, 2nd (hardound) + WebAssign Printed Access Card for Stewart's Essential Calculus: Early ... Multi-Term + WebAssign - Start Smart Guide

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