   Chapter 6.2, Problem 14QY ### 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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# In Exercises 9-14, use the integration table in Appendix C to find the indefinite integral. ∫ 2 x ( x + 1 ) e x 2 + 1 d x

To determine

To calculate: The solution of indefinite integral 2x(x2+1)ex2+1dx.

Explanation

Given Information:

The expression is provided as, 2x(x2+1)ex2+1dx.

Formula used:

The formula 37 for integral ueudu is,

ueudu=(u1)eu+C

General power differentiation Rule,

ddx[un]=nun1dudx

Calculation:

Let u=x2+1.

Differentiate the above equation with respect to x.

du=(2x+0)dx=2xdx

Consider the provided expression,

2x(x2+1)ex2+1dx

Rewrite.

2x(x2+1)ex2+1dx=(x2+1)ex2+12xdx

Substitute u for x2+1, and du for 2xdx

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