<s>[INST] Evaluate $\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$ 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)

 [/INST] 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
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