Explanation Given information: 2 θ 3 + 5 θ 2 + 8 θ + 4 ( θ 2 + 2 θ + 2 ) 2 Formula: ∫ 1 x d x = ln | x | + C ∫ 1 x 2 + 1 d x = tan − 1 x + C ∫ 1 x 2 d x = − 1 x + C Calculation: Express the integrand as a sum of partial fractions with undetermined coefficients. 2 θ 3 + 5 θ 2 + 8 θ + 4 ( θ 2 + 2 θ + 2 ) 2 = A θ + B ( θ 2 + 2 θ + 2 ) + C θ + D ( θ 2 + 2 θ + 2 ) 2 (1) Simplify the above equation. 2 θ 3 + 5 θ 2 + 8 θ + 4 ( θ 2 + 2 θ + 2 ) 2 = ( A θ + B ) ( θ 2 + 2 θ + 2 ) 2 + ( C θ + D ) ( θ 2 + 2 θ + 2 ) ( θ 2 + 2 θ + 2 ) ( θ 2 + 2 θ + 2 ) 2 2 θ 3 + 5 θ 2 + 8 θ + 4 = ( A θ + B ) ( θ 2 + 2 θ + 2 ) + ( C θ + D ) 2 θ 3 + 5 θ 2 + 8 θ + 4 = [ C ( θ ) + D + A ( θ 3 + 2 θ 2 + 2 θ ) + B ( θ 2 + 2 θ + 2 ) ] 2 θ 3 + 5 θ 2 + 8 θ + 4 = [ ( A ) θ 3 + ( 2 A + B ) θ 2 + ( 2 A + 2 B + C ) θ + D + 2 B ] (2) Equate the coefficient of the θ 2 term on both the sides of equation (2). 2 A + B = 5 (3) Equate the constant term on both the sides of equation (2). D + 2 B = 4 (4) Equate the coefficient of the θ 3 term on both the sides of equation (2). A = 2 Equate the coefficient of the θ term on both the sides of equation (2). 2 A + 2 B + C = 8 (5) Substitute 2 for A in equation (3)

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ISBN: 9780136208105

Author: Hass

Publisher: PEARSON

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Chapter 8.4, Problem 31E

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The integral through expansion of integrand as a sum of partial fractions.

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Chapter 8.4, Problem 30E

Chapter 8.4, Problem 32E

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