A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...

A Mind-blowing Solution by Cornel Ioan Valean

It is this way! Again, it is one of those integrals where it is incredibly difficult to imagine that a simple solution is possible. And yet, it is possible such a simple solution!

First, let the variable change $\displaystyle t\mapsto \frac{1-t}{1+t}$ in the main integral that becomes $$\int_0^1 \left(\frac{\pi}{4}+\arctan(t)\right)^2\frac{\log(t)}{1-t^2}\textrm{d}t=\underbrace{-\int_0^1 \frac{\displaystyle\left(\pi/2-\arctan(t)\right)^2 \operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle I}.$$

Now, the magical step is to consider the following result $\displaystyle\frac{(\pi/2-\arctan(t))^2}{t}=2\int_0^1 \frac{x\operatorname{arctanh}(x)}{t(t^2+x^2)}\textrm{d}x$ exploited in Cornel's answer here, and then we have that $$\small I=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}x\right)\textrm{d}t=-2\int_0^1 \left(\int_0^1\frac{x\operatorname{arctanh}(x)\operatorname{arctanh}(t)}{t(t^2+x^2)}\textrm{d}t\right)\textrm{d}x$$ $$=-2\left(\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t}\textrm{d}t}_{\displaystyle \pi^2/8}\right)^2+\underbrace{2\int_0^1 \left(\int_0^1\frac{t\operatorname{arctanh}(t)\operatorname{arctanh}(x)}{x(x^2+t^2)}\textrm{d}t\right)\textrm{d}x}_{\displaystyle -I},$$

whence, by symmetry reasons, we obtain the desired result $$\color{blue}{I=\int_0^1 \left(\frac{\pi}{4}+\arctan (t)\right)^2\frac{\log(t)}{1-t^2}\textrm{d}t=-\frac{\pi^4}{64}.}$$

End of story

Many such powerful auxiliary tools will be presented in the sequel of (Almost) Impossible Integrals, Sums, and Series.

A simple generalization of the main double integral

Based on the symmetry ideas above, there is room for generalizations of the type $$\color{red}{\int_a^b \left(\int_a^b\frac{x f(x)f(y)}{y(y^2+x^2)}\textrm{d}x\right)\textrm{d}y=\frac{1}{2}\left(\int_a^b\frac{f(x)}{x}\textrm{d}x\right)^2},$$ which may be a very strong result for deriving more difficult integrals.

A STUNNING GENERALIZATION (with the $n$th power of the inverse hyperbolic tangent) $$\color{purple}{\int_0^1 \frac{\operatorname{arctanh}^n(x)}{x}\Re\biggr\{\operatorname{Li}_{n+1}\left(\frac{1-x^2}{1+x^2}+i\frac{2 x}{1+x^2}\right)\biggr \}\textrm{d}x}$$ $$\color{purple}{=2^{-3 n-1} \left(2^{n+1}-1\right)n!\zeta^2(n+1)},$$ where it is used $\color{red}{\text{the generalization in red}}$ and the generalization at the point $i)$ in Cornel's post here - more similar results may be obtained with other of those generalizations, and some of them will appear in the sequel of (Almost) Impossible Integrals, Sums, and Series.

This not an answer

Considering that $$\big[\tan ^{-1}(t)\big]^2=\sum_{n=1}^\infty (-1)^{n+1}\,a_n\,t^{2n}$$ where $$a_n=\frac 1n\sum_{k=1}^n\frac 1 {2k-1}=\frac{H_{n-\frac{1}{2}}+2 \log (2)}{2 n}$$and using the fact that $$\int_0^1\frac {t^{2n}}{1-t^2}\log(t)=-\frac{1}{4} \zeta \left(2,\frac{2n+1}{2}\right)$$ $$K=2\int_0^1\big[\tan ^{-1}(t)\big]^2\,\frac{\log (t)}{1-t^2}\,dt$$ $$K=\frac 14 \sum_{n=1}^\infty (-1)^n \frac{\zeta \left(2,n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}+2 \log (2)\right)}{n}$$ which converges extremely slowly.