Đáp án A

Ta có \(f\left( {{x^2} + 4x} \right) = - 2{x^2} - 7x + 1 \Leftrightarrow \left( {2x + 4} \right)f\left( {{x^2} + 4x} \right) = \left( { - 2{x^2} - 7x + 1} \right)\left( {2x + 4} \right)\).

Lấy tích phân cận chạy từ 0 → 1 hai vế ta được:

\(\int\limits_0^1 {\left( {2x + 4} \right)} f\left( {{x^2} + 4x} \right)dx = \int\limits_0^1 {\left( { - 2{x^2} - 7x + 1} \right)\left( {2x + 4} \right)} dx = - \frac{{52}}{3}\).

Xét \(\int\limits_0^1 {\left( {2{\rm{x}} + 4} \right)f\left( {{x^2} + 4{\rm{x}}} \right)d{\rm{x}}} \). Đặt \(\left\{ {\begin{array}{*{20}{l}}{t = {x^2} + 4x \Rightarrow dt = \left( {2x + 4} \right)dx}\\{x = 0 \to t = 0,x = 1 \to t = 5}\end{array}} \right.\). Khi đó ta có:

\(\int\limits_0^1 {\left( {2x + 4} \right)} f\left( {{x^2} + 4x} \right)dx = \int\limits_0^5 {f\left( t \right)} dt = \int\limits_0^5 {f\left( x \right)} dx = - \frac{{52}}{3}\).

Xét \(I = \int\limits_0^5 {x.f'\left( x \right)dx} = \left. {xf\left( x \right)} \right|_0^5 - \int\limits_0^5 {f\left( x \right)} dx = - 40 - \left( { - \frac{{52}}{3}} \right) = - \frac{{68}}{3}\).