Calculus 8th Edition

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: 88

Answer

$-e^{-x}+2x+e^x+C$

Work Step by Step

$\int\frac{(1+e^x)^2}{e^x}dx$ $=\int\frac{1+2e^x+(e^x)^2}{e^x}dx$ $=\int\frac{1+2e^x+e^{2x}}{e^x}dx$ $=\int (\frac{1}{e^x}+\frac{2e^x}{e^x}+\frac{e^{2x}}{e^x})dx$ $=\int(e^{-x}+2+e^x)dx$ $=\int e^{-x}dx+\int(2+e^x)dx$ $=\int e^{-x}dx+(2x+e^x+C)$ Let $u=-x$. Then $du=-dx$, and $-du=dx$. $=\int e^u*(-1)du+2x+e^x+C$ $=-e^u+2x+e^x+C$ $=-e^{-x}+2x+e^x+C$
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