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: 24


$$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-\frac{x^{2}lnx}{2}+\frac{x^{2}}{4}+C$$

Work Step by Step

${(\frac{x^{2}ln^{2}x}{2})}'=x\,ln^{2}x+x\,lnx$ $$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-\int x\,lnx\,dx$$ ${(\frac{x^{2}lnx}{2})}'=x\,lnx+\frac{x}{2}$ $$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-[\frac{x^{2}lnx}{2}-\frac{1}{2}\int x\,dx]$$ $$=\frac{x^{2}ln^{2}x}{2}-\frac{x^{2}lnx}{2}+\frac{x^{2}}{4}+C$$
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