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 Notation

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}

= [- \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

\int_{- x - 2}^{x} x^{2} d x

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\int_{1}^{13} 6 x^{2} + 3 x - 2 d x

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