University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 8


$$\int x e^{3x}dx=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$$

Work Step by Step

$$A=\int x e^{3x}dx$$ Take $u=x$ and $dv=e^{3x}dx$ We then have $du=dx$ and $v=\frac{e^{3x}}{3}$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=\frac{xe^{3x}}{3}-\int\frac{e^{3x}}{3}dx$$ $$A=\frac{xe^{3x}}{3}-\frac{1}{3}\int e^{3x}dx$$ $$A=\frac{xe^{3x}}{3}-\frac{1}{3}\times\frac{e^{3x}}{3}+C$$ $$A=\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+C$$
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