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CalculusCalculus: An Applied Approach (MindTap Course List)10th Edition

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Chapter 5.2, Problem 32E
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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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.

∫ 4 x + 6 ( x 2 + 3 x + 7 ) 3   d x

To determine

To calculate: The indefinite integral of ∫4x+6(x2+3x+7)3dx and check the result by differentiating.

Explanation

Given Information:

The indefinite integral is ∫4x+6(x2+3x+7)3dx.

Formula used:

General Power Rule:

If u is a differentiable function of x, then

∫undudxdx=∫undu=un+1n+1+c,n≠−1

The Power Rule:

ddxxn=nxn−1

Where n is a real number.

Calculation:

Consider indefinite integral,

∫4x+6(x2+3x+7)3dx

The above indefinite integral ∫4x+6(x2+3x+7)3dx be written as ∫(x2+3x+7)−3(4x+6)dx.

Let u=x2+3x+7,

So,

du=(2x+3) dx

Now use the general power rule to get,

∫(x2+3x+7)−3(4x+6)dx=2∫(u)−3du

Now the integral will be:

2∫(u)−3du=2u−3+1−3+1

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