Define $\displaystyle f(x)=\int_{0}^{x}e^{-t^2}\mathrm{d}t$.
Evaluate $\displaystyle\int f(x)~\mathrm{d}x$. The final answer may involve $f(x)$. I know about $\displaystyle f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}f(x)=e^{-x^2}$, after that I can't solve it. Could you give me some hint for next step.
HINT: Use integration by parts on the integral: $$\int 1\cdot f(x) dx$$ with $dv=1$ and $u=f(x)$
Integration by parts: $$\int u \ dv = uv - \int v \ du$$