Đáp án:

\( 3 = \int_1^2 f(x-1) \, dx=\int_1^2 f(x-1) \, d(x-1) = \int_0^1 f(t) \, dt \)

\( \int_0^1 x^3 f'(x^2) \, dx = \frac{1}{2} \int_0^1 2x.x^2 f'(x^2) \, dx = \frac{1}{2} \int_0^1 t f'(t) \, dt \)

\( = \frac{1}{2} \left[ t f(t) \Big|_0^1 - \int_0^1 f(t) \, dt \right] = \frac{1}{2} \left[  f(1)  - \int_0^1 f(t) \, dt \right] \)

\( = \frac{1}{2} \left[ 4 - 3 \right] = \frac{1}{2} \Rightarrow \boxed{B}. \)