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: 9

Answer

$=\frac{1}{9}e^{3x}(3x-1)+C$

Work Step by Step

$\int xe^{3x}dx$ Use integration by parts once. $u=x, dv=\int e^{3x}dx$ $u'=dx, v=\frac{1}{3}e^{3x}$ $=uv-\int u'vdx$ $=x\frac{1}{3}e^{3x}-\int \frac{1}{3}e^{3x}dx$ $=\frac{1}{3}xe^{3x}-\frac{1}{9}e^{3x}+C$ $=\frac{1}{9}e^{3x}(3x-1)+C$
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