$t=\sqrt{1-{{x}^{2}}}\Rightarrow dt=-\frac{x}{\sqrt{1-{{x}^{2}}}}dx;{{x}^{2}}=1-{{t}^{2}}$

$\begin{align}  & \int{\frac{3{{x}^{3}}}{\sqrt{1-{{x}^{2}}}}}dx=\int{-3\left( 1-{{t}^{2}} \right)}dt=\int{\left( 3{{t}^{\text{3}}}-3 \right)}dt={{t}^{3}}-3t+C \\  & {{\left( \sqrt{1-{{x}^{2}}} \right)}^{3}}-3\sqrt{1-{{x}^{2}}}=\sqrt{1-{{x}^{2}}}\left( 1-{{x}^{2}}-3 \right)=-\left( {{x}^{2}}+2 \right)\sqrt{1-{{x}^{2}}} \\ \end{align}$

Chọn A