Question

id="TexFormula1" title=" \int\limits^5_1 { \frac{lnR}{ R^{2} } \, dR " alt=" \int\limits^5_1 { \frac{lnR}{ R^{2} } \, dR " align="absmiddle" class="latex-formula">

Answer 1

Integrate by parts, setting

$\begin{matrix}u=\ln R&&\mathrm dv=\dfrac{\mathrm dR}{R^2}\\\mathrm du=\dfrac{\mathrm dR}R&&v=-\dfrac1R\end{matrix}$

So the integral is

$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\dfrac{\ln R}R\bigg|_{R=1}^{R=5}+\int_1^5\frac{\mathrm dR}{R^2}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_{R=1}^{R=5}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=\frac15(4-\ln5)$