Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 17

Answer

$$\int\frac{e^u}{(1-e^u)^2}du=\frac{1}{1-e^u}+C$$

Work Step by Step

$$A=\int\frac{e^u}{(1-e^u)^2}du$$ $$A=\int e^u(1-e^u)^{-2}du$$ Let $r=1-e^u$. Then $dr=-e^udu$, so $e^udu=-dr$ Substitute into $A$, we have $$A=-\int r^{-2}dr$$ $$A=-\frac{r^{-1}}{-1}+C$$ $$A=\frac{1}{r}+C$$ $$A=\frac{1}{1-e^u}+C$$
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