Integration of Erf function Let $f(x) = \int^\sqrt{x}_1 e^{-t^2}dt$. Find $\int^1_0 \frac{f(x)}{
\sqrt{x}}dx$.
Could anyone give me any hint how to start? The Erf function $f(x)$ seems not to be easily integrated.

Brevan Ellefsen's solution is the most straightforward and efficient. Here is an alternative solution.
\begin{equation}
f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt
\label{eq:161108-1}
\tag{1}
\end{equation}
\begin{equation}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
\label{eq:161108-2}
\tag{2}
\end{equation}
We need the following result
\begin{equation}
\int \mathrm{erf}(x) dx = x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\label{eq:161108-3}
\tag{3}
\end{equation}
Proof: Integrate by parts
\begin{align}
\int \mathrm{erf}(x) dx &= x\,\mathrm{erf}(x) -\frac{2}{\sqrt{\pi}} \int x\,\mathrm{e}^{-x^{2}} dx \\
&= x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\end{align}
we used the substitution $u=x^{2}$.
Now we evaluate equation \eqref{eq:161108-1}
\begin{equation}
f(x) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(x) \Big|_{1}^{\sqrt{x}} = \frac{\sqrt{\pi}}{2} [\mathrm{erf}(\sqrt{x}) - \mathrm{erf}(1)]
\label{eq:161108-4}
\tag{4}
\end{equation}
Substitute equation \eqref{eq:161108-4} into equation \eqref{eq:161108-2} and evaluate the following two integrals.
\begin{align}
I_{1} &= \int\limits_{0}^{1} \frac{\mathrm{erf}(\sqrt{x})}{\sqrt{x}} dx \\
&= 2\int\limits_{0}^{1} \mathrm{erf}(z) dz \\
&= 2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}}
\end{align}
we used the substitution $z=\sqrt{x}$.
\begin{equation}
I_{2} = \int\limits_{0}^{1} \frac{1}{\sqrt{x}} dx = 2
\end{equation}
Putting all of the pieces together yields our final result
\begin{align}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
&= \frac{\sqrt{\pi}}{2} \left(2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}} - 2\,\mathrm{erf}(1) \right) \\
&= \frac{1}{\mathrm{e}} - 1
\end{align}
