Answer
$\frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$
Work Step by Step
$Expand$ $the$ $logarithmic$ $expression:$
$\ln \sqrt {\frac{x^2-1}{x^2+1}}$
Rewrite the root to exponent form
$$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}}$$
Use the Third Law of Logarithms
$$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} = \frac{1}{2}\ln (\frac{x^2-1}{x^2+1})$$
Use the Second Law of Logarithms [Distribute the $\frac{1}{2}$ as well]
$$\frac{1}{2}\ln (\frac{x^2-1}{x^2+1}) = \frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$$