Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 15

Answer

$\int\frac{e^x}{e^x-2e^{-2x}}dx$ = $\frac{1}{2}\ln(e^{2x}-2) + C$

Work Step by Step

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