Answer
$$ \displaystyle\int \sqrt{x}{\cdot}\sqrt{1-x}\, \mathrm d x = {{2\,\sqrt{1-x}\,x^{{{3}\over{2}}}-\sqrt{1-x}\,\sqrt{x}-\arctan \left({{\sqrt{1-x}}\over{\sqrt{x}}}\right)}\over{4}} $$
Explanation
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