1. Integral and limit

1 ) Calculate this integral :
( real and >2)
2 ) Calculate : .

2. What have you tried? Integration by parts and partial fraction decomposition will take care of the integral.

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

$\frac{1}{x}\frac{1}{x^2-1}=\frac{x}{x^2-1}-\frac{1}{x}$