## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 15

#### Answer

$$\frac{3}{8}$$

#### Work Step by Step

\eqalign{ & \int_e^{{e^2}} {\frac{{dx}}{{x{{\ln }^3}x}}} \cr & {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr & {\text{express the limits in terms of }}u \cr & x = e{\text{ implies }}u = \ln e = 1 \cr & x = {e^2}{\text{ implies }}u = \ln {e^2} = 2 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_e^{{e^2}} {\frac{{dx}}{{x{{\ln }^3}x}}} = \int_1^2 {\frac{1}{{{u^3}}}} du \cr & = \int_1^2 {{u^{ - 3}}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( { - \frac{{{u^{ - 2}}}}{2}} \right)} \right|_1^2 \cr & {\text{use the fundamental theorem}} \cr & = - \left( {\frac{{{{\left( 2 \right)}^{ - 2}}}}{2} - \frac{{{{\left( 1 \right)}^{ - 2}}}}{2}} \right) \cr & {\text{simplify}} \cr & = - \left( {\frac{1}{8} - \frac{1}{2}} \right) \cr & = \frac{3}{8} \cr}

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