How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s start with easy one
\begin{aligned}
I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G\end{aligned}

By my post
$$I_2= \frac{\pi}{4} \ln 2-G $$
and
$$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
$$
\begin{aligned}
I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G
\\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3})
\end{aligned}
$$
where the last integral refers to my post.

Let’s skip $I_5$ now.
$$
I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\
$$
$$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$
Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$

How far can we go with  the integral  $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$

We can proceed to $I_{8}$ now.
$$
\int_0^1 \frac{\ln \left(1-x^8\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} dx\\ \qquad\qquad =\frac{3 \pi}{4} \ln 2-2 G+ \int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} dx.
$$
In my post, two beautiful formula were found.
$$\boxed{\begin{align}
&\int_0^\infty \frac{\ln(1+x^{4m})}{1+x^2}dx
=2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{(2k-1)\pi}{8m}\bigg)\
\\
&\int_0^\infty \frac{\ln(1+x^{4m+2})}{1+x^2}dx
= \pi\ln2 + 2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{k\pi}{2(2m+1)}\bigg)
\end{align}}$$
As $$
\int_0^1 \frac{\ln \left(1+x^n\right)}{1+x^2} d x=\frac{1}{2}\left[\int_0^{\infty} \frac{\ln \left(1+x^n\right)}{1+x^2} d x-n G\right]
$$
Putting $m=1$ into the first formula in the box yields $$
\int_0^1 \frac{\ln \left(1+x^4\right)}{1+x^2} d x=\frac{1}{2}\left[2 \pi \ln \left(2 \cos \frac{\pi}{8}\right)-4 G\right]
$$
Hence
$$\boxed{\int_0^1 \frac{\ln \left(1-x^8\right)}{1+x^2} d x = \frac{3 \pi}{4} \ln 2-4 G+\pi \ln (\sqrt{2+\sqrt{2}}) }$$

Similarly, We can go further to $I_{12}$ now.
$$
\int_0^1 \frac{\ln \left(1-x^{12}\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x^6\right)}{1+x^2} dx\\ \qquad\qquad = \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G + \int_0^1 \frac{\ln \left(1+x^6\right)}{1+x^2} dx.
$$
Putting $m=1$ into the second formula in the box yields
$$\boxed{\int_0^1 \frac{\ln \left(1-x^{12}\right)}{1+x^2} d x= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G  + \frac{\pi}{2} \ln 6-3G= \frac{\pi}{4} \ln (72(7+4 \sqrt{3}))-6 G }$$
