## Calculus: Early Transcendentals 8th Edition

$\frac{-\ln 5 -1}{5} +1$
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 $R=5,1$ $$\frac{- \ln R -1}{R} \Bigg]^5_1= \frac{-\ln 5 -1}{5} +1$$