Answer
$\frac{1}{2}e^{x^{2}-1}+C$
Work Step by Step
$\int$$xe^{x^{2}-1}dx$
Let $u =$ $x^{2}-1$
$\frac{du}{dx}$ $=2x$
$dx=\frac{1}{2x}du$
$=$ $\int$$e^{u}du$
$=$ $e^{x^{2}-1} + C$
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 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 579: 2
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