Answer
$$\int (lnx)^{2}dx=x(lnx)^{2}-2xlnx+2x+C$$
Work Step by Step
${[x(lnx)^{2}]}'=(lnx)^{2}+2lnx$
$$\int (lnx)^{2}dx=x(lnx)^{2}-2\int lnx\,dx$$
${(x\,lnx)}'=lnx+1$
$$\int (lnx)^{2}dx=x(lnx)^{2}-2[xlnx-\int 1dx]$$
$$=x(lnx)^{2}-2xlnx+2x+C$$
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: 23
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