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

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

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

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

diverges

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

diverges

= diverges

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

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