Answer
$\frac{1}{2}[\ln{(x-1)}+\ln{(x+1)}-\ln{(x^2+1)}]$
Work Step by Step
Use the rule $\sqrt{a} = a^{\frac{1}{2}}$ to obtain:
$=\ln{\left(\dfrac{x^2-1}{x^2+1}\right)^{\frac{1}{2}}}$
RECALL:
(1) $\ln{(P^n)}=n \cdot \ln{P}$
(2) $\ln{(PQ)}=\ln{P} + \ln{Q}$
(3) $\ln{(\frac{P}{Q})} = \ln{P} - \ln{Q}$
Use rule (1) above to obtain
$=\frac{1}{2}\ln{\left(\dfrac{x^2-1}{x^2+1}\right)}$
Use rule (3) above to obtain:
$=\frac{1}{2}[\ln{(x^2-1)-\ln{(x^2+1)}}]$
Since $a^2-b^2=(a-b)(a+b)$, then the expression above is equivalent to:
$=\frac{1}{2}[\ln{[(x-1)(x+1)]}-\ln{(x^2+1)}]$
Use rule (2) above to obtain:
$=\frac{1}{2}[\ln{(x-1)}+\ln{(x+1)}-\ln{(x^2+1)}]$
