Lời giải chi tiết:

Ta có

\(\begin{array}{l}I = \int\limits_0^1 {\dfrac{{{x^2}{e^x}dx}}{{{{\left( {x + 2} \right)}^2}}}}  = \int\limits_0^1 {\left[ {\dfrac{{\left( {{x^2} - 4} \right){e^x}}}{{{{\left( {x + 2} \right)}^2}}} + \dfrac{{4{e^x}}}{{{{\left( {x + 2} \right)}^2}}}} \right]dx} \\ \Rightarrow I = \int\limits_0^1 {\left[ {\dfrac{{\left( {x - 2} \right){e^x}}}{{x + 2}} + \dfrac{{4{e^x}}}{{{{\left( {x + 2} \right)}^2}}}} \right]dx} \\ \Rightarrow I = \int\limits_0^1 {\left[ {\dfrac{{\left( {x - 2} \right){e^x}}}{{x + 2}} + \left( {\dfrac{{x - 2}}{{x + 2}}} \right)'.{e^x}} \right]dx} \\ \Rightarrow I = \int\limits_0^1 {\left( {\dfrac{{{e^x}\left( {x - 2} \right)}}{{x + 2}}} \right)} 'dx = \left. {\dfrac{{{e^x}\left( {x - 2} \right)}}{{x + 2}}} \right|_0^1 = \dfrac{{ - e}}{3} + 1 = \dfrac{{3 - e}}{3}\\ \Rightarrow \left\{ \begin{array}{l}a = 3\\b = 1\end{array} \right. \Rightarrow S = 2{a^2} + b = 19.\end{array}\)

Chọn A.