Đáp án:

\( (f(x))^3 + f(x) = x  \Rightarrow f'(x)(f(x))^3 + f'(x)f(x) = x f'(x) \)

\( \Rightarrow \int_0^2 f'(x)(f(x))^3 \, dx + \int_0^2 f'(x)f(x) \, dx = \int_0^2 x f'(x) \, dx \)

\( \Rightarrow \frac{(f(x))^4}{4} \big|_0^2 + \frac{(f(x))^2}{2} \Big|_0^2 = \int_0^2 x f'(x) \, dx \)

\( (f(0))^3 + f(0) = 0 \Rightarrow f(0)(1 + (f(0))^2) = 0 \Rightarrow f(0) = 0 \)

\( (f(2))^3 + f(2) - 2 = 0 \Rightarrow f(2) = 1 \)

\(\Rightarrow \int_0^2 x f'(x) \, dx = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \)

Đặt:  \( \begin{cases} 
u = x \\ 
dv = f'(x) \, dx 
\end{cases} 
\Rightarrow 
\begin{cases} 
du = dx \\ 
v = f(x) 
\end{cases}\)

\( \frac{3}{4} = \int_0^2 x f'(x) \, dx = x f(x)  \Big|_0^2 - \int_0^2 f(x) \, dx = 2 - \int_0^2 f(x) \, dx\)

\(\Rightarrow \int_0^2 f(x) \, dx = 2 - \frac{3}{4} = \frac{5}{4} \quad \Rightarrow \boxed{A}\)