   Chapter 6, Problem 30RE ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Using Integration Tables In Exercises 23–30, use the integration table in Appendix C to find the indefinite integral. ∫ 1 ( x 2 − 16 ) 2   d x

To determine

To calculate: The value of indefinite integral 1(x216)2dx.

Explanation

Given Information:

The provided expression is,

1(x216)2dx

Formula used:

The integral formula is,

1u2a2du=12aln|uau+a|+C

The integral formula is,

1(u2a2)ndu=12a2(n1)[u(u2a2)n1+(2n3)1(u2a2)n1du]

General power differentiation Rule:

ddx[un]=nun1dudx

Calculation:

Consider the provided expression

1(x216)2dx

Differentiate u=x with respect to x by the use of power rule of differentiation.

du=dx

Consider the provided expression,

1(x216)2dx

Rewrite the expression as.

1(x216)2dx=1(x242)2dx

Compare the expression with the integral formula,

1(u2a2)ndu

Here, a=4, n=2, u=x and du=dx.

Substitute u for x, a for 4, 2 for n and du for dx

Use the integral formula

1(u2a2)ndu=12a2(n1)[u(u2a2)n1+(2n3)1(u2a2)n1du]

And evaluate the provided integral

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