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

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

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

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

- x ln (1 - x^{2}) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x + C

Solution steps

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

Apply Integration By Parts

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

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

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

= - x ln (1 - x^{2}) - (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Simplify

= - x ln (1 - x^{2}) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x

Add a constant to the solution

= - x ln (1 - x^{2}) - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x + C

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