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

The integral of (ln(1+x))/(x^2) is -(ln(1+x))/x+ln|x|-ln|x+1|+C

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integral of (ln(1+x))/(x^2)

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

- \frac{ln ( 1 + x )}{x} + ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

\int \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

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (ln(1+x))/(x^2) ?
The integral of (ln(1+x))/(x^2) is -(ln(1+x))/x+ln|x|-ln|x+1|+C