What is the integral from 1 to 5 of (ln(R))/(R^2) ?

The integral from 1 to 5 of (ln(R))/(R^2) is (-ln(5)-1)/5+1

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integral from 1 to 5 of (ln(R))/(R^2)

\int_{1}^{5} \frac{ln ( R )}{R ^{2}} d R

\frac{- ln ( 5 ) - 1}{5} + 1

Decimal

0.47811 \dots

Solution steps

\int_{1}^{5} \frac{ln ( R )}{R ^{2}} d R

Apply Integration By Parts

= [- \frac{ln ( R )}{R} - \int - \frac{1}{R ^{2}} d R]_{1}^{5}

\int - \frac{1}{R ^{2}} d R = \frac{1}{R}

= [- \frac{ln ( R )}{R} - \frac{1}{R}]_{1}^{5}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= [\frac{- ln ( R ) - 1}{R}]_{1}^{5}

Compute the boundaries:

\frac{- ln ( 5 ) - 1}{5} + 1

= \frac{- ln ( 5 ) - 1}{5} + 1

What is the integral from 1 to 5 of (ln(R))/(R^2) ?
The integral from 1 to 5 of (ln(R))/(R^2) is (-ln(5)-1)/5+1