   Chapter 6.1, Problem 38E ### 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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# Finding an Indefinite Integral In Exercises 17-38, find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) ∫ x 3 e x 2 ( x 2 + 1 ) 2   d x

To determine

To calculate: The value of indefinite integral x3ex2(x2+1)2dx.

Explanation

Given Information:

The provided indefinite integral is x3ex2(x2+1)2dx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

eatdt=eata+C

Calculation:

Consider the indefinite integral x3ex2(x2+1)2dx

The above indefinite integral can be written as,

x3ex2(x2+1)2dx=(x2ex2)[x(x2+1)2]dx

Here,

dv=x(x2+1)2dx and u=x2ex2

First find v,

dv=x(x2+1)2dxdv=x(x2+1)2dx

On further solving,

v=12(x2+1) …...…... (1)

Now find du,

u=x2ex2

Differentiate both sides with respect x,

dudx=d(x2ex2)dxdudx=2xex2+x22xex2dudx=2xex2(x2+1)

And,

du=2xex2(x2+1)dx …...…..

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