Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$ I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$
My attempt: Letting
$$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$
$$
\begin{aligned}
I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta)
\end{aligned}
$$
$$
I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha
$$
But I can't go further.

The function does not have a elementary anti derivative.
Approximately (by numerical integration) the answer is $0.779611255707666$.
