Answer
$\frac{1}{3}ln(x+1)+\frac{1}{3}ln(x-2)-\frac{2}{3}ln(x+4)$
Work Step by Step
$ln[\frac{x^2-x-2}{(x+4)^2}]^{1/3}=\frac{1}{3}ln\frac{(x+1)(x-2)}{(x+4)^2}=\frac{1}{3}[ln(x+1)+ln(x-2)-2ln(x+4)]=\frac{1}{3}ln(x+1)+\frac{1}{3}ln(x-2)-\frac{2}{3}ln(x+4)$
