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}

= [- \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

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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