What is the integral of xln((1-x)/(1+x)) ?

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

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integral of xln((1-x)/(1+x))

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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