   Chapter 6.1, Problem 27E ### 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 ( ln   x ) 2 d x

To determine

To calculate: The value of indefinite integral x(lnx)2dx.

Explanation

Given Information:

The provided indefinite integral is x(lnx)2dx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

xndx=xn+1n+1,n1

Calculation:

Consider the indefinite integral x(lnx)2dx

Here,

dv=xdx and u=(lnx)2

First find v,

dv=xdxdv=xdx

On further solving,

v=x22 …...…... (1)

Now find du,

u=(lnx)2

Differentiate both side with respect x,

dudx=d((lnx)2)dxdudx=2lnxx

And,

du=2lnxxdx …...…... (2)

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

x(lnx)2dx=x22(lnx)2xlnxdx

Use integration by parts on the second integral xlnxdx

Here,

dv=xdx and u=lnx

First find v,

dv=xdxdv=xdx

On further solving,

v=x22 …...…... (3)

Now find du,

u=lnx

Differentiate both side with respect x,

dudx=d(lnx)dxdudx=1x

And,

du=1xdx

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