Answer
$\displaystyle \frac{4}{5}-\frac{\ln(5)}{5} ≈ 0.478112$
Work Step by Step
Integration by parts formula: $\displaystyle \int fg' $ $\displaystyle dx = fg-\int f'g$ $dx$
$\displaystyle \int_1^5\frac{\ln(R)}{R^2}dR = \int_1^5\ln(R)(\frac{1}{R^2})dR$
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Let $\displaystyle f = \ln(R)$ and $\displaystyle g' = \frac{1}{R^2}$
thus $\displaystyle f' = \frac{1}{R}$ and $\displaystyle g = -\frac{1}{R}$
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$\displaystyle \int_1^5\ln(R)(\frac{1}{R^2})dR = [\ln(R)(-\frac{1}{R})]_1^5-\int_1^5(-\frac{1}{R})(\frac{1}{R})dR$
$\displaystyle [-\frac{\ln(5)}{5}-(-\frac{\ln(1)}{1})] + \int_1^5\frac{1}{R^2}dR$
$\displaystyle -\frac{\ln(5)}{5} + 0 + (-\frac{1}{R})|^5_1$
$\displaystyle -\frac{\ln(5)}{5} + (-\frac{1}{5} - (-\frac{1}{1}))$
$\displaystyle -\frac{\ln(5)}{5} -\frac{1}{5} + 1$
$\displaystyle \frac{4}{5}-\frac{\ln(5)}{5} ≈ 0.478112$
