Im my recent answer https://math.stackexchange.com/a/4777055/198592 I found numerically that the following integral has a very simple result
$$i = \int_0^1 \frac{\text{Li}_2\left(\frac{i\; t}{\sqrt{1-t^2}}+1\right)+\text{Li}_2\left(1-\frac{i \;t}{\sqrt{1-t^2}}\right)}{\sqrt{1-t^2}} \, dt$$
Question: can you find it symbolically?
Here is what I did so far:
The substitution $t\to \sin(s)$ gives
$$i=\int_0^{\frac{\pi }{2}} (\text{Li}_2(i \tan (s)+1)+\text{Li}_2(1-i \tan (s))) \, ds$$
Now, using the relation
$$\text{Li}_2(z)= -\int_0^1 \frac{\log (1-x z)}{x} \, dx$$
we get
$$i= \int_0^{\frac{\pi }{2}} \int_0^{1}f(s,x)\, dsdx$$
where
$$f(s,x) = -\frac{1}{x}(\log (1-x (1+i \tan (s)))+\log (1-x (1-i \tan (s))))\\=- \frac{1}{x}\left(\log (1-x (1+i \tan (s)))(1-x (1-i \tan (s)))\right) \\= \frac{1}{x}\log \left(x^2 \tan ^2(s)+(x-1)^2\right) $$
Now it turns out that the integral over $s$
$$g(x) :=\int_0^{\frac{\pi }{2}} \log \left(x^2 \tan ^2(s)+(x-1)^2\right) \, ds$$
seems to vanishes for all $x \in(0,1)$.
Equivalently, with $s\to \tan ^{-1}(t)$ the integral is
$$g(x) = \int_{0}^{\infty} \frac{\log \left(t^2 x^2+(x-1)^2\right)}{t^2+1}\, dt$$
Here I am stuck at a similar problem as in the beginning.
