Answer
$2\ln x+\frac{1}{2}\ln ({1-x})$
Work Step by Step
We use the rules of logarithms to obtain:
$\ln \left( x^2\sqrt {1-x}\right) =\ln x^{2}+\ln \sqrt {1-x}=2\ln x+\ln \sqrt {1-x}$$=2\ln x+\frac{1}{2}\ln ({1-x})$
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