   Chapter 6.1, Problem 15E ### 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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# Integration by Parts In Exercises 5-16, use integration by parts to find the indefinite integral, See Examples 1, 2, 3, and 4. ∫ x 2 ( ln x ) 3   d x

To determine

To calculate: The value of indefinite integral x2(lnx)3dx.

Explanation

Given Information:

The provided indefinite integral is x2(lnx)3dx.

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 x2(lnx)3dx

Here,

dv=x2dx and u=(lnx)3

First find v,

dv=x2dxdv=x2dx

On further solving,

v=x33 …...…... (1)

Now find du,

u=(lnx)3

Differentiate both side with respect x,

dudx=d(lnx)3dxdudx=3(lnx)21xdudx=3(lnx)2x

And,

du=3(lnx)2xdx …...…... (2)

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

x2(lnx)3dx=13x3(lnx)313x33(lnx)21xdx=13x3(lnx)3x2(lnx)2dx

Use integration by parts on the second integral x2(lnx)2dx

Here,

dv=x2dx and u=(lnx)2

First find v,

dv=x2dxdv=x2dx

On further solving,

v=x33 …...…... (3)

Now find du,

u=(lnx)2

Differentiate both side with respect x,

dudx=d(lnx)2dxdudx=2(lnx)1xdudx=2lnxx

And,

du=2lnxxdx …...…... (4)

Apply integration by parts formula and substitute equation (3) and (4) in udv=uvvdu,

x2(lnx)3dx=13x3(lnx)3[13x3(lnx)213x3(2)(lnx)(1x)dx]=13x3(lnx)3[13x3(lnx)223x2(lnx)dx]=13x3(lnx)313x3(lnx)2+23x2(lnx)dx

Here,

dv=x2dx and u=(lnx)

First find v,

dv=x2dxdv=x2dx

On further solving,

v=x33

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