Answer
$\ln \left(\dfrac{1}{(x+1)^2}\right)$
Work Step by Step
We can write the expression as a single logarithm using the rules:
$\log_b(m)+\log_b(n)=\log_b(m\cdot n)$
and
$\log_b(m)+\log_b(n)=\log_b(\frac{m}{n})$
So,
$\ln \left(\dfrac{x-1}{x}\right)+\ln\left(\dfrac{x}{x+1}\right)-\ln(x^2-1)=$
$\ln \left(\dfrac{x-1}{x}\cdot\dfrac{x}{x+1}\div(x^2-1)\right)=$
$\ln \left(\dfrac{x-1}{x+1}\cdot\dfrac{1}{x^2-1}\right)=$
We can simplify:
$\ln \left(\dfrac{x-1}{x+1}\cdot\dfrac{1}{(x+1)(x-1)}\right)=$
$\ln \left(\dfrac{1}{(x+1)^2}\right)$
