   Chapter 6, Problem 8RE ### 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 1–8, use integration by parts to find the indefinite integral. ∫ ( ln   x ) 3   d x

To determine

To calculate: The infinite integral of (lnx)3dx by using the method of integration by parts.

Explanation

Given Information:

The provided integral is (lnx)3dx.

Formula used:

The method of integration by parts:

If v and u are two differentiable function of x. Then,

udv=uvvdu

Steps to solve the integral problems:

Step1: At first find the most complicated portion of the integrand and try to letter it as dv so that it can fit a fundamental integration rule. Then, the remaining factor or factors of the integrand will be u.

Step2: First find the factor whose derivative is simple and consider it as u and then the remaining factor or factors of the integrand will be dv and dv should always include the term dx of the original integrand.

Calculation:

Recall the provided integral.

(lnx)3dx

Observe from the above integrand that the simplest portion of the integrand is (lnx)3. So, consider, u=(lnx)3 and the remaining factors as dv=dx. Therefore,

du=3(lnx)21xdx

And,

dv=dx

So, integrate above expression as,

dv=dxv=x

Apply the integration by parts method.

udv=uvvdu

Substitute (lnx)3 for u, x for v, dx for dv, 3(lnx)21xdx for du and solve.

(lnx)3dx=(lnx)3x(x)3(lnx)21xdx=x(lnx)33(lnx)2dx

Now, consider the integrand 3(lnx)2dx. Here, the simplest portion of the integrand is (lnx)2. So, consider, u=(lnx)2 and the remaining factors as dv=3dx. Therefore,

du=2(lnx)1xdx

And,

dv=3dx

Integrate the above expression for v

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