Explanation Given: The integral function is ∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 d x Calculation: Show the integral function as follows: ∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 d x (1) Rearrange Equation (1). ∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 d x = ∫ x 2 + 1 ( x 2 − 2 x + 1 + 1 ) 2 d x = ∫ x 2 + 1 [ ( x − 1 ) 2 + 1 ] 2 d x (2) Consider x − 1 = tan θ (3) Differentiate both sides of the Equation (3). d x = sec 2 θ d θ Rearrange Equation (3). x = tan θ + 1 Substitute ( tan θ + 1 ) for x , ( tan θ ) for ( x − 1 ) , and ( sec 2 θ d θ ) for dx in Equation (2). ∫ x 2 + 1 [ ( x − 1 ) 2 + 1 ] 2 d x = ∫ ( tan θ + 1 ) 2 + 1 [ ( tan θ ) 2 + 1 ] 2 ( sec 2 θ d θ ) = ∫ tan 2 θ + 2 tan θ + 1 + 1 ( tan 2 θ + 1 ) 2 sec 2 θ d θ = ∫ tan 2 θ + 2 tan θ + 2 ( sec 2 θ ) 2 sec 2 θ d θ = ∫ tan 2 θ + 2 tan θ + 2 sec 2 θ d θ = ∫ ( tan 2 θ sec 2 θ + 2 tan θ sec 2 θ + 2 sec 2 θ ) d θ = ∫ ( sin 2 θ + 2 sin θ cos θ + 2 cos 2 θ ) d θ = ∫ [ 1 2 ( 1 − cos 2 θ ) + sin 2 θ + ( 1 + cos 2 θ ) ] d θ (4) Integrate Equation (4). ∫ [ 1 2 ( 1 − cos 2 θ ) + sin 2 θ + ( 1 + cos 2 θ ) ] d θ = [ 1 2 θ − 1 4 sin 2 θ − 1 2 cos 2 θ + θ + 1 2 sin 2 θ ] + C = [ 3 2 θ − 1 2 sin θ cos θ − 1 2 ( cos 2 θ − sin 2 θ ) + sin θ cos θ ] + C = [ 3 2 θ + 1 2 sin θ cos θ − 1 2 ( cos 2 θ − sin 2 θ ) ] + C = 1 2 [ 3 θ + sin θ cos θ − ( cos 2 θ − sin 2 θ ) ] + C (5) Rearrange Equation (3)

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ISBN: 9781133490975

Author: Stewart, James

Publisher: Cengage Learning

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Chapter 6.2, Problem 64E

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To evaluate: The integral function ∫x2+1(x2−2x+2)2dx.

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Chapter 6.2, Problem 63E

Chapter 6.2, Problem 59E

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Chapter 6 Solutions

Essential Calculus

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The integral function ∫ x 2 + 1 ( x 2 − 2 x + 2 ) 2 d x .

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To evaluate: The integral function ∫x2+1(x2−2x+2)2dx.

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Chapter 6 Solutions