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
Since the integrand is an even function, we can write
$$ I
= \frac{1}{2}\int_{-1}^{1} \log\left(\frac{1-x}{1+x}\right)\log\left(\frac{1-x^2}{1+x^2}\right)\,\frac{dx}{x}. $$
Now deforming the line contour $[-1, 1]$ to the semicircular contour from $-1$ to $1$ and substituting $x = e^{i\theta}$,
$$ I = -\frac{i}{2} \int_{0}^{\pi} \log(-i\tan(\theta/2)) \log(-i\tan \theta) \, d\theta, $$
where we utilized the identity $\frac{1-e^{i\theta}}{1+e^{i\theta}} = -i\tan(\theta/2)$. Now we note that, for $\theta \in (0, \pi/2) \cup (\pi/2, \pi)$,
$\log(-i\tan(\theta/2)) = \log\tan(\theta/2) - \frac{i\pi}{2}$,
$\log(-i\tan\theta) = \log\lvert\tan\theta\rvert - \operatorname{sign}(\tan\theta)\frac{i\pi}{2}$.
$\int_{0}^{\pi} \log\lvert\tan\theta\rvert \, d\theta = 2 \int_{0}^{\pi/2} (\log\sin\theta - \log\cos\theta) \, d\theta = 0$.
Plugging these back and taking real parts only (since we know that $I$ is real),
\begin{align*}
I
&= -\frac{\pi}{4} \int_{0}^{\pi} \left( \log\lvert\tan\theta\rvert + \operatorname{sign}(\tan\theta)\log\tan(\theta/2) \right) \, d\theta \\
&= -\frac{\pi}{2} \int_{0}^{\pi/2} \log\tan(\theta/2) \, d\theta \\
&= -\pi \int_{0}^{1} \frac{\log u}{1+u^2} \, du, \qquad (u=\tan(\theta/2)) \\
&= \pi C.
\end{align*}
Generalization. Utilizing a similar idea s in the computation above, we can prove that
Proposition. Let $p$, $q$ be positive integers. Write $g = \gcd(p,q)$ and assume that $p/g$ and $q/g$ are not simultaneously odd. Then
\begin{align*}
&\int_{0}^{1} \log\left(\frac{1-x^p}{1+x^p}\right)\log\left(\frac{1-x^q}{1+x^q}\right)\,\frac{dx}{x} \\
&\hspace{6em} = \pi \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \left( \frac{1}{p}\tan\left((2n+1)\frac{\pi p}{2q}\right) + \frac{1}{q} \tan\left((2n+1)\frac{\pi q}{2p}\right)\right)
\end{align*}
Of course, the above can be simplified further by using either Hurwitz zeta function or trigamma function
.
Best Answer
Alternatively, with $ \frac{1+t^2}{1-t^2+t^4}= \frac{1}{1+t^2} +\frac{3t^2}{1+t^6}$
\begin{align} &\int_0^1 \frac{(1+t^2)\ln t}{1-t^2+t^4}dt =\int_0^1 \underset{=-G}{\frac{\ln t}{1+t^2}}dt + \int_0^1 \underset{=-\frac13G}{\frac{3t^2\ln t}{1+t^6} \ \overset{t^3\to t}{dt}}=-\frac43G \end{align}