Answer
$$\int x\,lnx\,dx=\frac{x^{2}lnx}{2}-\frac{x^{2}}{4}+C$$
Work Step by Step
$${(\frac{x^{2}lnx}{2})}'=xlnx+\frac{x}{2}$$
$$\int x\,lnx\,dx=\frac{x^{2}lnx}{2}-\frac{1}{2}\int x\,dx$$
$$=\frac{x^{2}lnx}{2}-\frac{x^{2}}{4}+C$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 19
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