Đáp án đúng: D Giải chi tiết:Tích phân hai vế ta được : \(\begin{array}{l}\int\limits_0^1 {\left[ {f\left( x \right) + 4xf\left( {{x^2}} \right)} \right]dx} = \int\limits_0^1 {3xdx} \Rightarrow \int\limits_0^1 {f\left( x \right)dx} + 2\int\limits_0^1 {2xf\left( {{x^2}} \right)dx} = 3.\left. {\dfrac{{{x^2}}}{2}} \right|_0^1\\ \Rightarrow I + 2\int\limits_0^1 {2xf\left( {{x^2}} \right)dx} = \dfrac{3}{2}\end{array}\) Đặt \(t = {x^2} \Rightarrow dt = 2xdx\). Khi đó \(\int\limits_0^1 {2xf\left( {{x^2}} \right)dx} = \int\limits_0^1 {f\left( t \right)dt} = \int\limits_0^1 {f\left( x \right)dx} = I\). \( \Rightarrow I + 2I = \dfrac{3}{2} \Leftrightarrow I = \dfrac{1}{2}\). Vậy \(I = \dfrac{1}{2}\). Chọn D