Answer
$\frac{1}{2a}e^{ax^2}+C$
Work Step by Step
$\frac{dy}{dx}=xe^{ax^2}$
$dy=xe^{ax^2}dx$
$y=\int xe^{ax^2}dx$
let $u=ax^2$
$2ax$ $dx=du$
$y=\int xe^{ax^2}dx$
$=\frac{1}{2a}\int e^udu$
$=\frac{1}{2a}e^u+C$
$=\frac{1}{2a}e^{ax^2}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 121
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