Ta có:

$\begin{align}  & \int_{a}^{1}{\frac{2x}{{{\left( a-{{x}^{2}} \right)}^{2}}}dx}=\int_{a}^{1}{\frac{d\left( {{x}^{2}} \right)}{{{\left( a-{{x}^{2}} \right)}^{2}}}}=\int_{{{a}^{2}}}^{1}{\frac{dt}{{{\left( a-t \right)}^{2}}}}=\left. \frac{-1}{a-t} \right|_{{{a}^{2}}}^{1} \\  & \int_{a}^{1}{\frac{2x}{{{\left( a-{{x}^{2}} \right)}^{2}}}dx}=\frac{a+1}{a\left( 1-a \right)} \\ \end{align}$

Vậy đáp án đúng là C.