   Chapter 5.2, 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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# Applying the General Power Rule In Exercises 9–34, find the indefinite integral. Check your result by differentiating. See Examples 1, 2, 3, and 5. ∫ x 2 + 3 ( x 3 + 9 x − 4 ) 2 d x

To determine

To calculate: The indefinite integral of x2+3(x3+9x4)2dx and check the result by differentiating.

Explanation

Given Information:

The expression x2+3(x3+9x4)2dx.

Formula used:

General Power Rule:

If u is a differentiable function of x, then

undudxdx=undu=un+1n+1+C,n1

The Power Rule:

ddxxn=nxn1

Where n is a real number.

Calculation:

Consider indefinite integral,

x2+3(x3+9x4)2dx

This can be written as (x3+9x4)12(x2+3)dx.

Let u=x3+9x4,

So,

du=(3x2+9) dx=3(x2+3)dx

Now use the general power rule to get,

(x3+9x4)12(x2+3)dx=13(u)2du

Now the integral will be:

13(u)2du=13u2+12+1+C=13u11+C=13u1+C

Substitute back the value of u to get,

x2+3(x3+9x4)2dx=13(x3+9x4)1

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