   Chapter 6.1, Problem 37E ### 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 e 2 x ( 2 x + 1 ) 2   d x

To determine

To calculate: The value of indefinite integral xe2x(2x+1)2dx.

Explanation

Given Information:

The provided indefinite integral is xe2x(2x+1)2dx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

eaxdx=eaxa+c

Calculation:

Consider the indefinite integral xe2x(2x+1)2dx

Here,

dv=(2x+1)2dx and u=xe2x

First find v,

dv=(2x+1)2dx dv=(2x+1)2dx

On further solving,

v=12(2x+1) …...…... (1)

Now find du,

u=xe2x

Differentiate both side with respect x,

dudx=d(xe2x)dxdudx=2xe2x+e2xdudx=e2x(2x+1)

And,

du=e2x(2x+1)dx …...…... (2)

Apply integration by parts formula and substitute equation (1) and (2) in udv=uvvdu,

xe2x(2x+1)2dx=xe2x2(2x+1)+12e2xdx

Apply the formula eaxdx=eaxa+C,

xe2x(2x+1)2

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