Answer
$-\ln(1+e^{-x})+C$
Work Step by Step
$\int \frac{e^{-x}}{1+e^{-x}}dx$
let $1+e^{-x}=u$
$-e^{—x}dx=du$
$\int \frac{e^{-x}}{1+e^{-x}}dx$
$=-\int \frac{1}{u}du$
$=-\ln(U)+C$
$=-\ln(1+e^{-x})+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 99
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