The error function is defined as
$$\operatorname{erf}(x)=\frac 2 {\sqrt \pi}\int_{0}^x e^{-t^2}dt$$
It is not an elementary function. Since from the definition it is immediate (FTCI) that
$$\operatorname{erf}'(x)=\frac 2 {\sqrt \pi}e^{-x^2}$$
the primitive of $e^{-x^2}$ is expressible as
$$\int e^{-x^2} dx =\frac{\sqrt \pi}{2}\operatorname{erf}(x)-C $$
since any two primitives of a function $f$ differ by a constant (FTCII)
As a consequence your primitive can't be expressed in terms of elementary functions.
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With $\ds{r \equiv {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$
\begin{align}&\color{#c00000}{
\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x}
=\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over \pars{x - r}\pars{x - r^{*}}}\,\dd x
\\[3mm]&=\int_{0}^{1}\ln\pars{\ln\pars{1/x}}
\pars{{1 \over x - r} - {1 \over x - r^{*}}}\,{1 \over r - r^{*}}\,\dd x
\\[3mm] & =
{1 \over \Im\pars{r}}\,\Im\int_{0}^{1}
{\ln\pars{\ln\pars{1/x}} \over x - r}\,\dd x
\end{align}
With $\ds{x \equiv \expo{-t}}$:
\begin{align}&\color{#c00000}{
\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x}
={2\root{3} \over 3}\Im\int_{\infty}^{0}{\ln\pars{t} \over \expo{-t} - r}
\,\pars{-\expo{-t}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r}\int_{0}^{\infty}
{\ln\pars{t}\expo{-t} \over 1 - \expo{-t}/r}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\bracks{{1 \over r}
\sum_{n = 1}^{\infty}{1 \over r^{n - 1}}\int_{0}^{\infty}
\ln\pars{t}\expo{-nt}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\sum_{n = 1}^{\infty}r^{-n}\int_{0}^{\infty}t^{\mu}\expo{-nt}\,\dd t}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\sum_{n = 1}^{\infty}{r^{-n} \over n^{\mu + 1}}\Gamma\pars{\mu + 1}}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{r^{*}}}
\\[3mm]&=-\,{2\root{3} \over 3}\Im\lim_{\mu\ \to\ 0}\partiald{}{\mu}\bracks{
\Gamma\pars{\mu + 1}{\rm Li}_{\mu + 1}\pars{\expo{-\pi\ic/3}}}
\end{align}
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. Derivatives of the
PolyLogarithm, respect of the order, can be evaluated from its integral representation.
Also, see Hurwitz Zeta Function.
Best Answer
This integral appeared on an 1886 exam at the University of Cambridge and also discussed in A Treatise on the Integral Calculus by Joseph Edwards. In general we have
Proof :
\begin{align} \int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,\mathrm dx&=\int_0^1 \frac{\ln a+\ln t}{\sqrt{t}\;\sqrt{1-t}}\,\mathrm dt\tag1\\[7pt] &=\int_0^{\pi/2}\frac{\ln a+\ln\sin^2\theta }{\sqrt{\sin^2\theta}\;\sqrt{1-\sin^2\theta}}\cdot2\sin\theta\cos\theta\;\mathrm d\theta\tag2\\[7pt] &=2\ln a\int_0^{\pi/2}\;\mathrm d\theta+4\int_0^{\pi/2}\ln\sin\theta\;\mathrm d\theta\tag3\\[7pt] &=\pi\ln a-2\pi\ln2\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\pi\ln\left(\frac{a}{4}\right)}}\tag{$\color{red}{❤}$} \end{align} Hence $$\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}\,\mathrm dx=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large0}}$$
Explanation :
$(1)\;$ Use substitution $\;\displaystyle x=at$
$(2)\;$ Use substitution $\;\displaystyle t=\sin^2\theta\quad\implies\quad dt=2\sin\theta\cos\theta\;\mathrm d\theta$
$(3)\;$ Use Euler log-sine integral $\;\displaystyle \int_0^{\pi/2}\ln\sin\theta\;\mathrm d\theta=-\frac{\pi}{2}\ln2$