Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 2


$$\int xe^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C$$

Work Step by Step

$xe^{2x}={(\frac{xe^{2x}}{2})}'-\frac{e^{2x}}{2}$ $$\int xe^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{2}\int e^{2x}dx$$ $$=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C$$
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