Explanation Given Info: ∫ ( x + 1 ) x 2 + 4 x + 8 d x Formula Used: d d x ( x n ) = n x n − 1 ∫ x a d x = x a + 1 a + 1 + c Calculation: ∫ ( x + 1 ) x 2 + 4 x + 8 d x let u = x 2 + 4 x + 8 d u = 2 x + 4 d x then,we have ∫ ( x + 1 ) ( x 2 + 4 x + 8 ) 2 d x = 1 2 ∫ ( 2 x + 2 ) ( x 2 + 4 x + 8 ) 2 d x ∫ ( x + 1 ) ( x 2 + 4 x + 8 ) 2 d x = 1 2 ∫ ( 2 x + 2 + 2 − 2 ) ( x 2 + 4 x + 8 ) 2 d x ∫ ( x + 1 ) ( x 2 + 4 x + 8 ) 2 d x = 1 2 ∫ ( 2 x + 4 ) ( x 2 + 4 x + 8 ) 2 d x − ∫ 1 ( x 2 + 4 x + 8 ) 2 ∫ ( x + 1 ) ( x 2 + 4 x + 8 ) 2 d x = 1 2 ∫ 1 u 2 d u − ∫ 1 ( x 2 + 4 x + 8 ) 2 d x Now to evaluate the remaining integral , complete the square let w = x + 2 d w = d x ∫ 1 ( x 2 + 4 x + 8 ) 2 d x = ∫ 1 ( ( x + 2 ) 2 + 4 ) 2 d x ∫ 1 ( x 2 + 4 x + 8 ) 2 d x = ∫ 1 ( w 2 + 4 ) 2 d w Now , sub w=2tan θ dw=2sec 2 θ d θ w 2 + 4 = 4 tan 2 θ + 4 w 2 + 4 = 4 sec 2 θ ∫ 1 ( w 2 + 4 ) 2 d w = ∫ 2sec 2 θ d θ 16 sec 4 θ ∫ 1 ( w 2 + 4 ) 2 d w = 1 8 cos 2 θ d θ ∫ 1 ( w 2 + 4 ) 2 d w = 1 8 ( 1 2 θ + 1 2 sin θ ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 8.6, Problem 25E

To determine

To evaluate the integral $\int \frac{(x + 1)}{x^{2} + 4 x + 8} d x$

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Chapter 8.6, Problem 24E

Chapter 8.6, Problem 26E

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Chapter 8 Solutions

CALCULUS (CLOTH)

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