What is the integral from 1 to 2 of [(ln(x))/(x^2)] ?

The integral from 1 to 2 of [(ln(x))/(x^2)] is (-ln(2)-1)/2+1

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integral from 1 to 2 of [(ln(x))/(x^2)]

\int_{1}^{2} [\frac{ln ( x )}{x ^{2}}] d x

\frac{- ln ( 2 ) - 1}{2} + 1

Decimal

0.15342 \dots

Solution steps

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

Apply Integration By Parts

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

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

= [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{2}

Since the denominators are equal, combine the fractions:

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

= [\frac{- ln ( x ) - 1}{x}]_{1}^{2}

Compute the boundaries:

\frac{- ln ( 2 ) - 1}{2} + 1

= \frac{- ln ( 2 ) - 1}{2} + 1

What is the integral from 1 to 2 of [(ln(x))/(x^2)] ?
The integral from 1 to 2 of [(ln(x))/(x^2)] is (-ln(2)-1)/2+1