Answer
$ln5+2lnx+\frac{1}{3}ln(1-x)-ln4-2ln(x+1)$
Work Step by Step
$ln\frac{5x^2\ \sqrt[3] {1-x}}{4(x+1)^2}=ln\frac{5x^2\ (1-x)^{1/3}}{4(x+1)^2}=ln5+2lnx+\frac{1}{3}ln(1-x)-ln4-2ln(x+1)$
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