$I=\int\limits_{0}^{1}{{{(x-1)}^{3}}\sqrt{2x-{{x}^{2}}}dx}=\int\limits_{0}^{1}{({{x}^{2}}-2x+1)\sqrt{2x-{{x}^{2}}}(x-1)dx}$

Đặt $t=\sqrt{2x-{{x}^{2}}}\Rightarrow {{t}^{2}}=2x-{{x}^{2}}\Rightarrow tdt=(1-x)dx.t(0)=0;t(1)=1$

$I=\int\limits_{0}^{1}{(1-{{t}^{2}})t(-t)dt=\int\limits_{0}^{1}{({{t}^{4}}-{{t}^{2}})dt}}=\left( \frac{{{t}^{5}}}{5}-\frac{{{t}^{3}}}{3} \right)\left| \begin{matrix} 1  \\ 0  \\ \end{matrix} \right.=\frac{1}{5}-\frac{1}{3}=-\frac{2}{15}$

Vậy đáp án đúng là B.