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$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm erf}\pars{x}\ln\pars{x}
\,\dd x}$. Let's
$\ds{{\cal I}\pars{\mu}
\equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm erf}\pars{x}\,\dd x}$ such that $\ds{I = \lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}}$
Since
$\ds{{\rm erf}\pars{x}
\stackrel{{\rm def.}}{=}{2 \over \root{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$:
\begin{align}
{\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}}
{2 \over \root{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x
\\[3mm]&=
{2 \over \root{\pi}}\int_{0}^{\pi/2}\dd\theta\,\cos^{\mu}\pars{\theta}
\Theta\pars{\cos\pars{\theta} - \sin\pars{\theta}}
\overbrace{\int_{0}^{\infty}\dd r\,r^{\mu + 1}\expo{-r^{2}}}
^{\Gamma\pars{1 + \mu/2}/2}
\\[3mm]&={1 \over \root{\pi}}\,\Gamma\pars{1 + {\mu \over 2}}
\int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta}
\end{align}
\begin{align}
I&=\lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}=
{1 \over 2\root{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\
\overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}}
\\[3mm]&+
{1 \over \root{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}}
\overbrace{%
\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}}
^{\ds{\braces{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}}
\end{align}
$\Gamma\pars{z}$ and $\Psi\pars{z}$ are the $\it Gamma$ and $\it Digamma$ functions, respectively. $\gamma$ and $G$ are the $\it Euler-Mascheroni$ and Catalan constants, respectively.
$$
\begin{array}{|l|}\hline \mbox{}\\
\quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}}
\,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad
\\[2mm] =
\quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi -
2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over
{\displaystyle 16\root{\pi}}}\quad
\\ \mbox{}\\ \hline
\end{array}
\approx 0.0436462
$$
Rewrite
$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$
Then plugging this in and reversing the order of integration, we get the integral; value as
$$\begin{align}\int_0^1 \frac{du}{u^2} \, \int_0^{\infty} dx \, \left (\frac1{\frac1{u^2}+x^2} \frac1{1+x^2} \right ) &= \int_0^1 \frac{du}{1-u^2} \, \int_0^{\infty} dx \left ( \frac1{1+x^2}-\frac1{\frac1{u^2}+x^2} \right )\\ &= \int_0^1 \frac{du}{1-u^2} \, \frac{\pi}{2} (1-u) \\ &= \frac{\pi}{2} \log{2}\end{align}$$
Best Answer
$$I= \int_0^{\infty} \frac{ \operatorname{erf}^2(x)}{x^2}dx=\int_0^{\infty} \operatorname{erf}^2(x)d\left(\frac{-1}x\right)$$ Integration by part
$$I=-\frac{\operatorname{erf}^2(x)}x\bigg|_0^\infty+ \frac{4}{\pi}\int_0^{\infty}\frac{1}x 2\operatorname{erf}(x)\cdot e^{-x^2}dx=\frac{8}{\pi}\int_0^{\infty}\int_0^{x}\frac{1}x e^{-t^2-x^2}dtdx$$
Use polar coordinate $x=r\cos\theta, t=r\sin\theta$:
$$\begin{align}I&=\frac{8}{\pi}\int_0^{\pi/4}\int_0^{\infty}\frac{1}{r\cos\theta} e^{-r^2}r~drd\theta\\ \\ &=\frac{8}{\pi}\int_0^{\pi/4}\frac{1}{\cos\theta}d\theta\int_0^{\infty} e^{-r^2}~dr\\ \\ &=\frac{8}{\pi}\cdot\ln(1+\sqrt2)\cdot\frac{\sqrt\pi}2\\ \\ &=\frac{4}{\sqrt\pi}\ln(1+\sqrt2)\end{align}$$