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

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

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

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

diverges

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{x ( 1 + x )} d x = - ln ∣ x ∣ + ln ∣ x + 1 ∣

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

Simplify

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

Compute the boundaries:

diverges

= diverges

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