Answer
$- e^{-x}+\dfrac{4^x}{\ln 4} +C$
Work Step by Step
Given: $\int (e^{-x} +4^x) dx$
Thus, $\int (e^{-x} +4^x) dx=(-1) e^{-x}+\dfrac{4^x}{\ln 4} +C=- e^{-x}+\dfrac{4^x}{\ln 4} +C$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 53
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