Integrate by parts, setting
[tex]\begin{matrix}u=\ln R&&\mathrm dv=\dfrac{\mathrm dR}{R^2}\\\mathrm du=\dfrac{\mathrm dR}R&&v=-\dfrac1R\end{matrix}[/tex]
So the integral is
[tex]\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}[/tex]
[tex]\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_{R=1}^{R=5}[/tex]
[tex]\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)[/tex]
[tex]\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=\frac15(4-\ln5)[/tex]
