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: 23


$$\int (lnx)^{2}dx=x(lnx)^{2}-2xlnx+2x+C$$

Work Step by Step

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