Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 91

Answer

$\frac{1}{1-e^u}+C$

Work Step by Step

Let $v=1-e^u$. Then $dv=-e^u du$, and $-dv=e^u du$. $\int\frac{e^u}{(1-e^u)^2} du$ $=\int \frac{-dv}{v^2}$ $=-\int v^{-2}dv$ $=-\frac{v^{-1}}{-1}+C$ $=v^{-1}+C$ $=\frac{1}{v}+C$ $=\frac{1}{1-e^u}+C$
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