Answer
$3e^{x^{1/3}}+c $
Work Step by Step
Recall that $(e^x)'=e^x$
Let $ u= x^{1/3}$, then $ du=\frac{1}{3} x^{-2/3}dx $ and hence we have
$$
\int x^{-2/3}e^{x^{1/3}} dx=3\int e^{u} du\\
=3e^u+c=3e^{x^{1/3}}+c
.
$$
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: 86
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