Answer is: option2
\( -\frac{1}{6} \)Solution:
\[ f(x)=\int_{0}^{x}(1+t)e^{-t}\,dt \]
\[ f(0)=0 \]
\[ f'(x)=(1+x)e^{-x} \]
\[ f''(x)=e^{-x}-(1+x)e^{-x}=-xe^{-x} \]
\[ f'''(x)=-e^{-x}+xe^{-x} \]
\[ f'(0)=1,\quad f''(0)=0,\quad f'''(0)=-1 \]
Therefore, \[ f(x)=0+1x+0\frac{x^2}{2!}+(-1)\frac{x^3}{3!}+\cdots =x-\frac{x^3}{6}+\cdots \]
The coefficient of \(x^3\) is \[ -\frac{1}{6} \]
