Answer
$ e^{x} -\frac{1}{3}e^{3x}+c $
Work Step by Step
Recall that $(e^x)'=e^x$
We have
$$
\int \frac{e^{2x}-e^{4x}}{e^x}dx= \int e^{ x}-e^{3x}dx=e^{x} -\frac{1}{3}e^{3x}+c
.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 328: 81
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