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

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

Popular >

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

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}^{b}

Compute the boundaries:

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

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

\int_{0}^{1} (x e^{- x^{2}}) d x

\int_{0}^{1} 40 x (2 x^{2} + 1)^{4} d x

\int_{π}^{2 π} cos^{2} (x) d x

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

\int_{0}^{8} 64 - x^{2} d x

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