Question $1$

$2$ pts

If $F(x)={\displaystyle \underset{x}{\overset{-1}{}}}{e}^{t+t^2}dt,$ then

$\frac{d}{dx}F(x)=e^{x+x^2}$

$\frac{d}{dx}F(x)=-e^{x+x^2}(1+2x)$

$\frac{d}{dx}F(x)=-e^{x+x^2}$

$\frac{d}{dx}F(x)=-e^{x+x^2}+C$