Explanation Calculation: Simplify the term 3 x 2 + 4 x − 6 x 2 − 3 x + 2 in the given integrand by using long division method, x 2 − 3 x + 2 3 3 x 2 + 4 x − 6 3 x 2 − 9 x + 6 13 x − 12 ¯ Thus, ∫ 3 x 2 + 4 x − 6 x 2 − 3 x + 2 d x = ∫ ( 3 + 13 x − 12 x 2 − 3 x + 2 ) d x (1) Simplify the function 13 x − 12 x 2 − 3 x + 2 by using method of partial fraction, 13 x − 12 x 2 − 3 x + 2 = 13 x − 12 ( x − 1 ) ( x − 2 ) = A x − 1 + B x − 2 Thus, 13 x − 12 x 2 − 3 x + 2 = A x − 1 + B x − 2 (2) Multiply both sides of equation (1) by ( x − 1 ) ( x − 2 ) , A ( x − 2 ) + B ( x − 1 ) = 13 x − 12 (3) Substitute x = 1 in equation (3), A ( 1 − 2 ) + B ( 1 − 1 ) = 13 ( 1 ) − 12 − A = 1 A = − 1 Thus the value of A = − 1

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 7.5, Problem 65E

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To evaluate: The integral is ∫3x2+4x−6x2−3x+2dx.

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Chapter 7.5, Problem 64E

Chapter 7.5, Problem 66E

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Chapter 7 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The integral is ∫ 3 x 2 + 4 x − 6 x 2 − 3 x + 2 d x .

Solution Summary: The author evaluates the value of the integral by using a long division method.

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To evaluate: The integral is ∫3x2+4x−6x2−3x+2dx.

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Chapter 7 Solutions