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Evaluate the integral.

$ \displaystyle \int_1^5 \frac{\ln R}{R^2} dR $

$\int_{1}^{5} \frac{\ln R}{R^{2}} d R=\left[-\frac{1}{R} \ln R\right]_{1}^{5}-\int_{1}^{5}-\frac{1}{R^{2}} d R=-\frac{1}{5} \ln 5-0-\left[\frac{1}{R}\right]_{1}^{5}=-\frac{1}{5} \ln 5-\left(\frac{1}{5}-1\right)=\frac{4}{5}-\frac{1}{5} \ln 5$

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