Integrating by parts twice,
$$ \begin{align} \int_{0}^{1} (\arctan x)^{2} \ dx &= x (\arctan x)^{2} \Big|^{1}_{0} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - 2 \int_{0}^{1} \frac{x \arctan x}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \arctan(x) \ln(1+x^{2}) \Big|^{1}_{0} + \int_{0}^{1} \frac{\ln (1+x^{2})}{1+x^{2}} \ dx \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 + \int_{0}^{1} \frac{\ln(1+x^{2})}{1+x^{2}} \ dx \end{align}$$
Let $x = \tan t $.
Then
$$\begin{align}\int_{0}^{1} (\arctan x)^{2} \ dx &=\frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \int_{0}^{\pi /4} \ln (\cos t) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 -2 \int_{0}^{\pi /4} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2nt) - \ln 2 \right) \ dt \\ &= \frac{\pi^{2}}{16} - \frac{\pi}{4} \ln 2 - 2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_{0}^{\pi /4} \cos (2nt) \ dt + \frac{\pi}{2} \ln 2 \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=1}^{\infty} \frac{\sin \left(\frac{\pi n}{2} \right)}{n^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\ &= \frac{\pi^{2}}{16} + \frac{\pi}{4} \ln 2 - C \end{align}$$
Fourier series of Log sine and Log cos
We have the following closed form evaluation.
$$
I:=\int_0^{\Large \frac{\pi}{4}}\!\!\left(\frac{1}{\ln(\tan x)}+\frac{1}{1-\tan x}\right)\! \mathrm dx= \color{blue}{\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\!\left(\!\frac14\!\right)} \tag1
$$
where $\gamma$ is the Euler-Mascheroni constant.
A numerical approximation is
$$
I =\color{blue}{0.462999316582640135993449151416572....}
$$
As Lucian pointed out, by the change of variable $\displaystyle u=\tan x$, we readily have
$$
I=\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{1}{1+u^2} \mathrm du.
$$
We set
$$
I(s):=\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \mathrm du, \quad s>-1.
$$
The expected integral is thus $I(0)$.
Let's differentiate $I(s)$.
We get
$$
\begin{align}
I'(s)& =\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s\ln u}{1+u^2} \mathrm du \\\\
& =\int_0^1\left(1+\frac{\ln u}{1-u}\right)\frac{u^s}{1+u^2} \mathrm du \\\\
& =\int_0^1\! \frac{u^s}{1+u^2}\mathrm du +\int_0^1\!\frac{(1+u)u^s \ln u}{(1-u^2)(1+u^2)}\mathrm du \\\\
& =\int_0^1\! \frac{u^s(1-u^2)}{1-u^4}\mathrm du +\int_0^1\!\frac{u^s \ln u}{1-u^4}\mathrm du+\int_0^1\!\frac{u^{s+1} \ln u}{1-u^4}\mathrm du.
\end{align}
$$
By the change of variable $\displaystyle v=u^4$ in each of the preceding integrals and recalling the well known integral representations for the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$ and for its derivative,
$$
\psi(s) = -\gamma+\int_0^1 \frac{1 - v^{s-1}}{1 -v}{\rm{d}} v, \quad s>0,
$$
$$
\psi'(s) = -\int_0^1 \frac{s^{s-1} \ln v}{1 - v}{\rm{d}} v, \quad s>0,
$$
we obtain
$$
I'(s)=\frac{1}{4}\psi\left(\frac{s+3}{4}\right)-\frac{1}{4}\psi\left(\frac{s+1}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+2}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+1}{4}\right). \tag2
$$
Since $$ \left|\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \right| < u^s, \quad 0<u<1,\, s>-1,$$
giving
$$ |I(s)|\leq \int_0^1\left|\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \right|\mathrm du < \!\!\int_0^1 u^s\mathrm du = \frac{1}{s+1}, $$
then, as $s \rightarrow +\infty$, we have $I(s) \rightarrow 0$.
We deduce that
$$
I(s)=\log \Gamma \left(\frac{s+3}{4}\right)\!-\!\log \Gamma \left(\frac{s+1}{4}\right) \!-\!\frac{1}{4}\psi\!\left(\frac{s+2}{4}\right)\!-\!\frac{1}{4}\psi\!\left(\frac{s+1}{4}\right)
, \, s>-1, \tag3
$$
and, for $s=0$,
$$
\begin{align}
I=\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\left(\frac14\right)
\end{align}
$$
where have used special values of the digamma function,
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac14\right) & = -\gamma + \frac\pi2- 3\ln 2,
\end{align}
$$
and the complement/reflection formula
$$
\Gamma\left(\frac34\right)\Gamma\left(\frac14\right)=\frac{\pi}{\sin(\frac\pi4)}=\pi \sqrt{2}.
$$
Best Answer
Using the Fourier series of $\ln(\tan{x})$, \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\ =&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\ &\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\ =&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\ =&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G} \end{align}
Things could be made clearer if we explicitly write out the terms of the sums. For the red sums, \begin{align} &-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\ =&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\ =&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G \end{align}