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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

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