Answer
$\frac{1}{1-e^u}+C$
Work Step by Step
Let $v=1-e^u$. Then $dv=-e^u du$, and $-dv=e^u du$.
$\int\frac{e^u}{(1-e^u)^2} du$
$=\int \frac{-dv}{v^2}$
$=-\int v^{-2}dv$
$=-\frac{v^{-1}}{-1}+C$
$=v^{-1}+C$
$=\frac{1}{v}+C$
$=\frac{1}{1-e^u}+C$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 454: 91
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