Answer
$\int(-xe^{\frac{x^2}{2}})dx=-e^{\frac{x^2}{2}}+C$
Work Step by Step
Substitution:
$u=\frac{x^2}{2}$
$\frac{du}{dx}=x$
$dx=\frac{1}{x}du$
$\int(-xe^{\frac{x^2}{2}})dx=\int(-xe^u)\frac{1}{x}du=\int-e^udu=-e^u+C=-e^{\frac{x^2}{2}}+C$
Finite Math and Applied Calculus (6th Edition)
by Waner, Stefan; Costenoble, Steven
Published by
Brooks Cole
ISBN 10:
1133607705
ISBN 13:
978-1-13360-770-0
Chapter 13 - Review - Review Exercises - Page 1004: 15
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