What is the integral from 0 to 1 of (ln(x))/(x^2) ?

The integral from 0 to 1 of (ln(x))/(x^2) is diverges

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integral from 0 to 1 of (ln(x))/(x^2)

\int_{0}^{1} \frac{ln ( x )}{x ^{2}} d x

diverges

Solution steps

\int_{0}^{1} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{0}^{1}

= [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{0}^{1}

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

= [- \frac{ln ( x )}{x} - \frac{1}{x}]_{0}^{1}

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}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int sin (x^{2}) + cos (2 x) - \frac{1}{2} d x

\int_{0}^{2} \frac{x}{x ^{2} + 6 x + 13} d x

(1 + 5 x)^{9}

\int \frac{9}{25 + 16 x ^{2}} d x

(sin (2 x))^{^{'}}

What is the integral from 0 to 1 of (ln(x))/(x^2) ?
The integral from 0 to 1 of (ln(x))/(x^2) is diverges