Answer
$$\frac{1}{2(e^{-x}+2)^2}+c$$
Work Step by Step
Let $u=e^{-x}+2$, and then $du=-e^{-x}dx$, hence we have
$$\int \frac{e^{-x}dx}{(e^{-x}+2)^3}=-\int \frac{du}{u^3}=-\frac{1}{-2}u^{-2}+c\\
=\frac{1}{2(e^{-x}+2)^2}+c$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 83
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