Answer
$-2 \ln(x-1)$
Work Step by Step
Recall:
$\log_a (MN) = \log_a M+\log_a N$
Use the rule above to obtain:
\begin{align*}\require{cancel}
\ln \left(\dfrac{x}{x-1} \right) + \ln \left(\dfrac{x+1}{x} \right) -\ln{(x^2-1)}\\\\
&= \ln \left(\dfrac{(x)(x+1)}{(x-1)(x)} \right)-\ln{(x^2-1)}\\\\
&= \ln \left(\dfrac{\cancel{(x)}(x+1)}{(x-1)\cancel{(x)}} \right)-\ln{(x^2-1)}\\\\
&=\ln{\left(\dfrac{x+1}{x-1}\right)}-\ln{(x^2-1)}
\end{align*}
Recall further that:
$\log_a\left(\dfrac{M}{N}\right) = \log_a M-\log_a N$
Use this rule to obtain:
$\ln \left(\dfrac{x+1}{x-1} \right)- \ln(x^2-1) = \ln \left(\dfrac{x+1}{(x-1)(x^2-1)} \right)$
Since $x^2-1=(x+1)(x-1)$, then
\begin{align*}
\require{cance}
\ln \left(\dfrac{x+1}{(x-1)(x^2-1)} \right) \\\\
&= \ln \left(\dfrac{x+1}{(x-1)(x-1)(x+1)} \right) \\\\
&= \ln \left(\dfrac{\cancel{x+1}}{(x-1)(x-1)\cancel{(x+1)}} \right) \\\\
&= \ln \left(\dfrac{1}{(x-1)^2} \right)
\end{align*}
With $\dfrac{1}{(x-1)^2} = (x-1)^2$, then:
$\ln \left(\dfrac{1}{(x-1)^2} \right) = \ln{(x-1)^{-2}}$
Recall that:
$\log_a M^r = r \log_a M$
Using the rule above gives:
$\ln((x-1)^{-2} = -2 \ln(x-1)$
Therefore,
$\ln \left(\dfrac{x}{x-1} \right) + \ln \left(\dfrac{x+1}{x} \right) - \ln(x^2-1) = \boxed{-2 \ln(x-1)}$
