Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 522: 13

Answer

$\left( {\ln y} \right)\ln \left( {\ln y} \right) - \left( {\ln y} \right) + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{\ln \left( {\ln y} \right)}}{y}} dy \cr & = \int {\ln \left( {\ln y} \right)\left( {\frac{1}{y}} \right)} dy \cr & {\text{Let }}t = \ln y,{\text{ }}dt = \frac{1}{y}dy \cr & {\text{Substituting}} \cr & \int {\ln \left( {\ln y} \right)\left( {\frac{1}{y}} \right)} dy = \int {\ln t} dt \cr & {\text{Integrating by parts}} \cr & {\text{Let }}u = \ln t,{\text{ }}du = \frac{1}{t}dt \cr & dv = dt,{\text{ }}v = t \cr & {\text{Using integration by parts formula}} \cr & \int {\ln t} dt = t\ln t - \int {t\left( {\frac{1}{t}} \right)} dt \cr & = t\ln t - \int {dt} \cr & = t\ln t - t + C \cr & {\text{Write in terms of }}y,{\text{ substitute }}\ln y{\text{ for }}t \cr & = \left( {\ln y} \right)\ln \left( {\ln y} \right) - \left( {\ln y} \right) + C \cr} $$
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