Evaluate $\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$

Question

Question

Evaluate $$\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$$

Here is my work (and where I stopped)

Solution found on Internet

(I didn’t understand it at all to be honest)

Answer 1

You should get $Li_{2}(x)$ function

https://mathworld.wolfram.com/Dilogarithm.html

$\int{\frac{ln(x)-x}{x^2-1}dx} = \int{\frac{ln{(x)}}{x^2-1}}-\int{\frac{x}{x^2-1}dx}\\ =-\int{\frac{ln{(x)}}{1-x^2}}dx - \frac{1}{2}\int{\frac{2x}{x^2-1}dx}\\ =-\int{\frac{ln{(x)}}{(1-x)(1+x)}}dx - \frac{1}{2}\ln{|x^2-1|} \\ -\frac{1}{2}\int{\frac{2ln{(x)}}{(1-x)(1+x)}dx} - \frac{1}{2}\ln{|x^2-1|}\\ -\frac{1}{2}\int{\frac{ln{(x)}((1-x)+(1+x))}{(1-x)(1+x)}dx} - \frac{1}{2}\ln{|x^2-1|}\\ -\frac{1}{2}\left(\int{\frac{ln{(x)}}{1+x}dx}+\int{\frac{ln{(x)}}{1-x}dx}\right) - \frac{1}{2}\ln{|x^2-1|}\\ -\frac{1}{2}\left(\ln(x+1)\ln(x) - \int{\frac{\ln(x+1)}{x}dx}+\int{\frac{\ln{(x)}}{1-x}dx}\right) - \frac{1}{2}\ln{|x^2-1|} $

This two integrals can be expressed using $Li_{2}(x)$ function
For integral $\int{\frac{ln{(1+x)}}{x}dx}$ we can use substitution $u = -x $
For integral $\int{\frac{ln{(x)}}{1-x}dx}$ we can use substitution $v = 1-x$ and we will get $Li_{2}(x)$ function