Đáp án:

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

Xét \( I = \int_0^1 x^3 f'(x^2) dx \)

\( \begin{cases} u = x^2 \\ dv = x f'(x^2) dx \end{cases} \Rightarrow \begin{cases} du = 2x dx \\  v = \frac{1}{2} f(x^2)\end{cases} \)

\(\left( v = \int x f'(x^2) dx = \frac{1}{2} \int f'(x^2) d(x^2) = \frac{1}{2} f(x^2) \right)\)

\(I = \frac{1}{2} x^2 f(x^2) \big|_0^1 - \frac{1}{2} \int_0^1 2x f(x^2) dx\)

\(= 2 - \frac{1}{2} \int_0^1 f(x^2) dx^2 = 2 - \frac{1}{2} \int_0^1 f(t) dt = 2 - \frac{3}{2} = \frac{1}{2} \Rightarrow \boxed{B} \)

Gợi ý: \( \int f'(u) du = f(u)\)