What is the integral from 1 to 3 of (ln(x))/(x^2) ?

The integral from 1 to 3 of (ln(x))/(x^2) is (-ln(3)-1)/3+1

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integral from 1 to 3 of (ln(x))/(x^2)

\int_{1}^{3} \frac{ln ( x )}{x ^{2}} d x

\frac{- ln ( 3 ) - 1}{3} + 1

Decimal

0.30046 \dots

Solution steps

\int_{1}^{3} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{3}

= [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{3}

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

= [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{3}

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}^{3}

Compute the boundaries:

\frac{- ln ( 3 ) - 1}{3} + 1

= \frac{- ln ( 3 ) - 1}{3} + 1

\int_{0}^{3} 5 x - 8 d x

\int_{5}^{6} x^{2} + x - 14 d x

\int_{0}^{2} x^{\frac{1}{2}} d x

\int_{- 2}^{0} - 5 e^{- 7 x + 5} d x

\int_{0}^{5} \frac{x ^{3}}{1 + x ^{2}} d x

What is the integral from 1 to 3 of (ln(x))/(x^2) ?
The integral from 1 to 3 of (ln(x))/(x^2) is (-ln(3)-1)/3+1