Với F(x) = $A\sqrt{1-{{x}^{3}}}+\frac{B}{1+\sqrt{x}}$ thì

$F'(x)=\frac{A.-3{{x}^{2}}}{2\sqrt{1-{{x}^{3}}}}-\frac{B\frac{1}{2\sqrt{x}}}{{{\left( 1+\sqrt{x} \right)}^{2}}}=\frac{-3A}{2}.\frac{{{x}^{2}}}{\sqrt{1-{{x}^{3}}}}+\frac{-B}{2}\frac{1}{\sqrt{x}{{\left( 1+\sqrt{x} \right)}^{2}}}$

Lại có:

${{F}^{'}}(x)=f(x)\Rightarrow \frac{-3A}{2}.\frac{{{x}^{2}}}{\sqrt{1-{{x}^{3}}}}+\frac{-B}{2}.\frac{1}{\sqrt{x}{{\left( 1+\sqrt{x} \right)}^{2}}}=\frac{{{x}^{2}}}{\sqrt{1-{{x}^{3}}}}+\frac{1}{\sqrt{x}{{\left( 1+\sqrt{x} \right)}^{2}}}$

$\Rightarrow \left\{ \begin{align} & \frac{-3A}{2}=1 \\ & \frac{-B}{2}=1 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & A=\frac{-2}{3} \\ & B=-2 \\ \end{align} \right.\Rightarrow A+B=\frac{-8}{3}$

Đáp án đúng là B