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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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In Exercises 16–21, use integration by parts or the integration table in Appendix C to evaluate the definite integral.

1 e ( ln 6 x ) 2 d x

To determine

To calculate: The value of definite integral 1e(ln6x)2dx.

Explanation

Given Information:

The provided definite integral is,

1e(ln6x)2dx

Formula used:

(1) The formula 44 for integral (lnu)2du is:

(lnu)2du=u[22lnu+(lnu)2]+C

(2) General power differentiation Rule:

ddx[un]=nun1dudx

Calculation:

Consider u=6x.

Differentiate the considered function with respect to x using power rule of differentiation.

du=6dx

Consider the provided integral:

1e(ln6x)2dx

Multiply and divide by 6.

1e(ln6x)2dx=161e6(ln6x)2dx=161e(ln6x)26dx

Substitute u for x and du for 6dx.

1e(ln6x)2dx=161e(lnu)2du

Use the formula 44 and solve the above integral as:

1e(ln6x)2dx=16[u[22lnu+(lnu)2]]1e

Substitute 6x for u

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