Calculus (3rd Edition)

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: 21


$$\int x^{2}lnx\,dx=\frac{x^{3}lnx}{3}-\frac{x^{3}}{9}+C$$

Work Step by Step

$${(\frac{x^{3}lnx}{3})}'=x^{2}lnx+\frac{x^{2}}{3}$$ $$\int x^{2}lnx\,dx=\frac{x^{3}lnx}{3}-\frac{1}{3}\int x^{2}dx$$ $$=\frac{x^{3}lnx}{3}-\frac{x^{3}}{9}+C$$
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