Find $\int x \sqrt{1-x}dx$

Let $1-x=u^2$. Then $x=1-u^2$ and $dx=-2udu$. So
\[\int x\sqrt{1-x}dx=(1-u^2) \sqrt{u^2}\cdot (-2udu)=-2\int u^2(1-u^2)du\]
\[=-2\int 1-u^4du=-2(u-\frac{u^5}{5})+c=-2(\sqrt{1-x}-\frac{(\sqrt{1-x})^5}{5})+c\]