Ta có \(F\left( x \right) = \int\limits_{}^{} {{e^{{x^2}}}\left( {{x^3} - 4x} \right)dx}  = \int\limits_{}^{} {{e^{{x^2}}}\left( {{x^2} - 4} \right)xdx} \)

Đặt \(t = {x^2} \Rightarrow dt = 2xdx \Rightarrow F\left( t \right) = \frac{1}{2}\int\limits_{}^{} {{e^t}\left( {t - 4} \right)dt} \).

Đặt \(\left\{ \begin{array}{l}u = t - 4\\dv = {e^t}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = {e^t}\end{array} \right.\)

\( \Rightarrow F\left( t \right) = \frac{1}{2}\left[ {\left( {t - 4} \right){e^t} - \int\limits_{}^{} {{e^t}dt} } \right] = \frac{1}{2}\left[ {\left( {t - 4} \right){e^t} - {e^t}} \right] = \frac{1}{2}\left( {t - 5} \right){e^t} + C.\)

\(\begin{array}{l} \Rightarrow F\left( x \right) = \frac{1}{2}\left( {{x^2} - 5} \right){e^{{x^2}}} + C \Rightarrow g\left( x \right) = F\left( {{x^2} + x} \right) = \frac{1}{2}\left[ {{{\left( {{x^2} + x} \right)}^2} - 5} \right]{e^{{{\left( {{x^2} + x} \right)}^2}}} + C\\ \Rightarrow g'\left( x \right) = \frac{1}{2}\left[ {2\left( {{x^2} + x} \right)\left( {2x + 1} \right){e^{{{\left( {{x^2} + x} \right)}^2}}} + \left( {{{\left( {{x^2} + x} \right)}^2} - 5} \right){e^{{{\left( {{x^2} + x} \right)}^2}}}.2\left( {{x^2} + x} \right).\left( {2x + 1} \right)} \right]\\\,\,\,\,\,\,g'\left( x \right) = \left( {{x^2} + x} \right)\left( {2x + 1} \right){e^{{{\left( {{x^2} + x} \right)}^2}}}\left( {{{\left( {{x^2} + x} \right)}^2} - 4} \right)\\\,\,\,\,\,\,g'\left( x \right) = x\left( {x + 1} \right)\left( {2x + 1} \right)\left( {{x^2} + x - 2} \right)\left( {{x^2} + x + 2} \right){e^{{{\left( {{x^2} + x} \right)}^2}}}\\g'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x =  \pm 1\\x = \frac{{ - 1}}{2}\\x =  - 2\end{array} \right.\end{array}\)

Vậy hàm số \(F\left( {{x^2} + x} \right)\) có 5 điểm cực trị.

Chọn B.