Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$
Taking integration by parts,
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx = -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$
Now
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx = \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$
and the first integral is easily computed by integration by parts:
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx = -\log 2 + \frac{\pi}{2}. $$
The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion
\begin{align*} \log \sec\theta = -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right| &= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\ &= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta), \end{align*}
we have
\begin{align*} \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx &= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\ &= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,} \end{align*}
where $G$ is Catalan's constant. Combining altogether, we obtain
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx = \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$
Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields
