Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 1

Answer

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

Work Step by Step

$\int xe^{2x}$ $u = x$ $u' = 1$ $\frac{du}{dx} = 1$ $du = 1 dx$ $du = dx$ $dv = e^{2x}$ $v = \frac{1}{2}e^{2x}$ $uv - \int vdu$ $= (x)(\frac{1}{2}e^{2x}) - \int \frac{1}{2}e^{2x}dx$ $= (x)(\frac{1}{2}e^{2x}) - \frac{1}{2}\times\frac{1}{4}e^{2x}$ $= (x)(\frac{1}{2}e^{2x}) - \frac{1}{4}e^{2x}$ $= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x}$ $= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C$
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