Chapter 7.5, Problem 59E

### Single Variable Calculus

8th Edition
James Stewart
ISBN: 9781305266636

Chapter
Section

### Single Variable Calculus

8th Edition
James Stewart
ISBN: 9781305266636
Textbook Problem

# Evaluate the integral.59. ∫ d x x 4 − 16

To determine

To evaluate: The integral function 1x416dx.

Explanation

Given information:

The integral function is âˆ«1x4âˆ’16â€‰dx.

Calculation:

Show the integral function as follows:

âˆ«1x4âˆ’16â€‰dx (1)

Modify Equation (1) as shown below:

âˆ«1x4âˆ’16â€‰dx=âˆ«1(x2)2âˆ’42â€‰dx==âˆ«1(x2+4)(x2âˆ’4)â€‰dx=âˆ«1(x2+4)(x2âˆ’22)â€‰dx=âˆ«1(x2+4)(x+2)(xâˆ’2)â€‰dx (2)

Consider the function y=1(x2+4)(x+2)(xâˆ’2) (3)

Expand Equation (3) using partial fraction decomposition.

1(x2+4)(x+2)(xâˆ’2)=Ax+B(x2+4)+C(x+2)+D(xâˆ’2)

1=(Ax+B)(x+2)(xâˆ’2)+C(x2+4)(xâˆ’2)+D(x2+4)(x+2) (4)

Equate the coefficients to get the value of A, B, C, and D.

Equate the coefficients of x3 in Equation (4).

A+C+D=0 (5)

Substitute 2 for x in Equation (4).

1=(Ax+B)(x+2)(xâˆ’2)+C(x2+4)(xâˆ’2)+D(x2+4)(x+2)=(2A+B)(2+2)(2âˆ’2)+C(22+4)(2âˆ’2)+D(22+4)(2+2)=D(4+4)Ã—41=32DD=132 (6)

Substitute âˆ’2 for x in Equation (4).

1=(Ax+B)(x+2)(xâˆ’2)+C(x2+4)(xâˆ’2)+D(x2+4)(x+2)=(âˆ’2A+B)(âˆ’2+2)(âˆ’2âˆ’2)+C((âˆ’2)2+4)(âˆ’2âˆ’2)+D((âˆ’2)2+4)(âˆ’2+2)=C(4+4)Ã—(âˆ’4)1=âˆ’32CC=âˆ’132 (7)

Modify Equation (5) using (6) and (7).

Substitute (âˆ’132) for C and (132) for D in Equation (5)

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