Answer
$ -2ln(x+1)$
Work Step by Step
$ln(\frac{x-1}{x})+ln(\frac{x}{x+1})-ln(x^2-1)=ln(\frac{x-1}{x}\cdot \frac{x}{x+1})-ln((x+1)(x-1))=ln(\frac{x-1}{x+1})-ln((x+1)(x-1))=ln(\frac{x-1}{x+1}\cdot \frac{1}{(x+1)(x-1)})=ln\frac{1}{(x+1)^2}=-2ln(x+1)$
where $x\ne 0,\pm1$
