Lời giải

\( \int_0^1 f(x) \, dx = \int_0^1 6x^2 f(x^3) \, dx - \int_0^1 \frac{6}{\sqrt{3x+1}} \, dx \).  

\( \int_0^1 f(x) \, dx = 2 \int_0^1 f(t) \, dt - 4\sqrt{3x+1}\Big|_0^1 \) 

\( \left(t = x^3 \quad vì  \quad (\sqrt{3x+1})' =  \frac{3}{2\sqrt{3x+1}} \right) \)

\( \Rightarrow \int_0^1 f(x) \, dx =4\sqrt{3x+1}\Big|_0^1= 4 \Rightarrow \boxed{B} \)