Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 507: 22


$$\int \frac{lnx}{x\sqrt{1+(lnx)^{2}}}dx=\sqrt{1+(lnx)^{2}}+C$$

Work Step by Step

$$let\,t=1+(lnx)^{2},\,dt=\frac{2lnx}{x}dx$$ $$\int \frac{lnx}{x\sqrt{1+(lnx)^{2}}}dx=\int \frac{1}{2}t^{-\frac{1}{2}}dt=\sqrt{t}+C$$ $$=\sqrt{1+(lnx)^{2}}+C$$
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