Answer
$\ln\sqrt {\frac{(x+3)^2x}{x^2-1}}$
Work Step by Step
$\frac{1}{2}[2\ln(x+3)+\ln x-\ln(x^2-1)]=\frac{1}{2}[\ln(x+3)^2+\ln x-\ln(x^2-1)]=\frac{1}{2}(\ln\frac{(x+3)^2x}{x^2-1})=\ln[\frac{(x+3)^2x}{x^2-1}]^{\frac{1}{2}}=\ln\sqrt {\frac{(x+3)^2x}{x^2-1}}$
