What is the integral from 1 to 4 of (ln(x))/(x^2) ?

The integral from 1 to 4 of (ln(x))/(x^2) is (-2ln(2)-1)/4+1

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integral from 1 to 4 of (ln(x))/(x^2)

\int_{1}^{4} \frac{ln ( x )}{x ^{2}} d x

\frac{- 2 ln ( 2 ) - 1}{4} + 1

Decimal

0.40342 \dots

Solution steps

\int_{1}^{4} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

= [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

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

= [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{4}

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}^{4}

Compute the boundaries:

\frac{- 2 ln ( 2 ) - 1}{4} + 1

= \frac{- 2 ln ( 2 ) - 1}{4} + 1

What is the integral from 1 to 4 of (ln(x))/(x^2) ?
The integral from 1 to 4 of (ln(x))/(x^2) is (-2ln(2)-1)/4+1