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

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

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

Simplify

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

= \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

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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