Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 104



Work Step by Step

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