Question: Is there any particular inequality involving $e^{x^2}$ that I need to use here? In any case, I couldn't solve it.

Answer: 
$$\int\_{0}^{x}e^{x^2-t^2}\,dt \stackrel{t=xs}{=} x\int\_{0}^{1}e^{x^2(1-s^2)}\,ds > x\int\_{0}^{1}1\,ds = x.$$
Set $e^{-t^2}=u$ and $dt=dv$ we have
$$\int\_{0}^{x}e^{-t^2}dt=xe^{-x^2}+2\int\_{0}^{x}t^2e^{-t^2}dt$$
since $x>0$ we have
$$\int\_{0}^{x}e^{-t^2}dt> xe^{-x^2}$$