Calculus (3rd Edition)

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



Work Step by Step

Let $u=e^{-x}+2$, and then $du=-e^{-x}dx$, hence we have $$\int \frac{e^{-x}dx}{(e^{-x}+2)^3}=-\int \frac{du}{u^3}=-\frac{1}{-2}u^{-2}+c\\ =\frac{1}{2(e^{-x}+2)^2}+c$$
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