Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises: 5

Answer

$-\frac{1}{3} te^{-3t}-\frac{1}{9}e^{-3t}+C$

Work Step by Step

Let $u=t$ and $dv=e^{-3t} dt$ then $du = dt$ and $v=-\frac{1}{3}e^{-3t}$ and therefore $\int te^{-3t}dt = -\frac{1}{3}te^{-3t}-\int -\frac{1}{3}e^{-3t}dt=-\frac{1}{3}te^{-3t}+\frac{1}{3}\int e^{-3t}dt=-\frac{1}{3} te^{-3t}-\frac{1}{9}e^{-3t}+C$
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