Answer
$e^{-x}-\frac{5}{2e^{2x}}+C$
Work Step by Step
$\int \frac{5-e^x}{e^{2x}}dx$
$=-\int e^{-2x}(e^x-5)dx$
$=-(\int e^{-x}dx - \int 5e^{-2x}dx$
$=-(-e^{-x}+\frac{5}{2}e^{-2x})+C$
$=e^{-x}-\frac{5}{2e^{2x}}+C$
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: 105
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