   Chapter 6, Problem 2TYS ### 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 1–3, use integration by parts to find the indefinite integral. ∫ x 2 ln   9 x   d x

To determine

To calculate: The indefinite integral of x2ln9xdx by using the method of integration by parts.

Explanation

Given Information:

The integral is x2ln9xdx.

Formula used:

The integration by part of two differentiable function is,

udv=uvvdu

The basic rule of integration for constant C is,

Cdx=0

The simple power rule of integration is,

xndx=xn+1n+1+C

Where, n1.

The chain rule of differentiation is,

ddx(f(g))=(f(g))ddx(g)

Calculation:

Consider the provided integral, x2ln9xdx.

Consider the first function is ln9x so,

u=ln9x

Differentiate both side with respect to x, use the chain rule of differentiation.

du=19x(9)dx=99xdx=1xdx

Consider the second function x2 so,

dv=x2dx

Integrate both side with respect to x, use the simple power rule of integration.

dv=x2dxv=13x3

Now, use the integration by parts method,

udv=uvvdu

Substitute ln9x for u, 13x3 for v, x2dx for dv, 1xdx for du and solve

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