Let $h:[a,b]\to\mathbb{R}$ be differentiable on $(a,b)$. If $h'(x)>0$ for all $x\in(a,b)$ then $h$ is strictly increasing.

We know that $h'(x)>0$ so we must have
\[
h'(c) = \frac{h(x)-h(y)}{x-y} > 0
\]
If $x<y$, then $x-y<0$ and so we must have $h(x)-h(y)<0\Rightarrow h(x)<h(y)$.
If $x>y$, then $x-y>0$ and so we must have $h(x)-h(y)>0\Rightarrow h(x)>h(y)$.

Thus $h$ is increasing.

I ommitted some things which are required for a rigorous proof, but the idea should be clear. Since 
\[
h(x)=\int_0^x e^{t^2}\mathrm{d}t 
\]
we have
\[
h'(x) = e^{x^2} > 0\Rightarrow  h\textrm{ is increasing.}