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: 
**Hint**. One may differentiate
$$
f(x)=\int\_{0}^{x}e^{-t^2}\,dt-xe^{-x^2} 
$$ obtaining
$$
f'(x)=2x^2e^{-x^2}>0,\quad x>0,
$$ then one may use $f(0)=0$.
The rectangle $[0,x]\times[0,e^{-x^2}]$ lies entirely below the graph of $e^{-x^2}$.