Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 27


$\frac{-\ln 5 -1}{5} +1$

Work Step by Step

First, solve for the indefinite integral: let $u=\ln R$ and $dv=R^{-2} \ dR$. Then $du= \frac{1}{R}$ and $v=-\frac {1}{R}$. Apply integration by parts. $$\int \frac{\ln R}{R^2}= \frac{-\ln R}{R}$- \int \frac{-1}{R^2} \ dR$$ $$=\frac{-\ln R}{R} - \frac{1}{R} = \frac{- \ln R -1}{R}$$ Evaluate at $R=5,1$ $$ \frac{- \ln R -1}{R} \Bigg]^5_1= \frac{-\ln 5 -1}{5} +1$$
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