#### Answer

$-2 \ln (x+1)$

#### Work Step by Step

Use rule $\ln_a{(mn)}=\log_a{m} + \log_a{n}$ to obtain:
$\ln [(\dfrac{x-1}{x})(\dfrac{x}{x+1})(\dfrac{1}{x^2-1})] =\ln [\dfrac{1}{(x+1)^2}]$
Now, use $\log_a (\dfrac{m}{n})=\ln m -\ln n$ to obtain:
$\ln [\dfrac{1}{(x+1)^2}]=\ln (1)-\ln (x+1)^2$
Finally, use $\log_a{a^m}=m\log_a{m}$ to obtain:
$\ln (1)-\ln (x+1)^2=-2 \ln (x+1)$