Calculus 10th Edition

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: 2

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$ $=$ $e^{x^{2}-1} + C$
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