Answer
$y=\frac{1}{2}e^{2x}-2x-\frac{1}{2}e^{-2x}+C$
Work Step by Step
$\frac{dy}{dx}=(e^x-e^{-x})^2$
$\frac{dy}{dx}=(e^{2x}-2+e^{-2x})$
$dy=(e^{2x}-2+e^{-2x})dx$
$y=\int(e^{2x}-2+e^{-2x})dx$
$y=\frac{1}{2}e^{2x}-2x-\frac{1}{2}e^{-2x}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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: 122
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