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  $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
 A: The odd cases are much harder to evaluate. Below are some of them:
\begin{align}
\int_0^1 \frac{\ln(1-x^5)}{1+x^2}dx= & \ \frac{9\pi}8\ln2-\frac{13}5G+\frac\pi2\ln\left(\cos\frac\pi{20}\cos\frac{3\pi}{20} \right)\\
&\>\>\>+ \frac{3\pi}{20}\ln \tan\frac{3\pi}{20} 
 - \frac{\pi}{20}\ln \tan\frac{\pi}{20} \\
\\
\int_0^1 \frac{\ln(1-x^7)}{1+x^2}dx = & \ \frac{13\pi}8\ln2-\frac{24}7G+\frac\pi2\ln\left(\cos\frac\pi{28}\cos\frac{3\pi}{28} \cos\frac{5\pi}{28}\right)\\
&\>\>\>+  \frac{5\pi}{28}\ln \tan\frac{5\pi}{28}-\frac{3\pi}{28}\ln \tan\frac{3\pi}{28} 
 +\frac{\pi}{28}\ln \tan\frac{\pi}{28} \\
\end{align}
A: For even cases, apply
\begin{align}
&1-x^{4m}=(1-x^4) \prod_{k=1}^{m-1} \left(1+2x^2\cos\frac{k\pi}{m}+x^4 \right)\\
& 1-x^{4m+2}= (1-x^2)\prod_{k=0}^{m-1} \left(1+2x^2\cos\frac{(2k+1)\pi}{2m+1}+x^4\right)
\end{align}
and
$$\int_0^1 \frac{\ln(1+2x^2\cos \theta +x^4)}{1+x^2}dx
=\pi \ln\left(2\cos\frac{\theta}4\right)-2G
$$
to obtain
\begin{align}
&\int_0^1\frac{\ln(1-x^{2n})}{1+x^2}dx
=-nG+\frac{(2n-1)\pi}4\ln2+\pi \sum_{k=1}^{[\frac{n-1}2]}\ln  \cos\frac{(n-2k)\pi}{4n}
\end{align}
In particular
\begin{align}
 \int_0^1\frac{\ln(1-x^{2})}{1+x^2}dx
=& -G+\frac{\pi}4\ln2 \\
\int_0^1\frac{\ln(1-x^{4})}{1+x^2}dx
=& -2G+\frac{3\pi}4\ln2 \\
\int_0^1\frac{\ln(1-x^{6})}{1+x^2}dx
=& -3G-\frac{\pi}4\ln2 +\pi \ln(1+\sqrt3)\\
 \int_0^1\frac{\ln(1-x^{8})}{1+x^2}dx
=& -4G+\frac{3\pi}4\ln2 +\frac\pi2\ln(2+\sqrt2)\\
 \int_0^1\frac{\ln(1-x^{10})}{1+x^2}dx 
=&-5G-\frac{3\pi}4\ln2 +\pi \ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\
\int_0^1\frac{\ln(1-x^{12})}{1+x^2}dx
=& -6 G+\frac{\pi}4\ln2 +\pi\ln(3+\sqrt3)\\
 \int_0^1\frac{\ln(1-x^{14})}{1+x^2}dx 
=& -7G+ \frac{13\pi}4\ln2 +\pi \ln\left(\cos\frac\pi{28} \cos\frac{3\pi}{28} \cos\frac{5\pi}{28} \right)\\
 \int_0^1\frac{\ln(1-x^{16})}{1+x^2}dx 
=&-8G+\frac{5\pi}4\ln2 +\pi\ln\left(1+\sqrt2+\sqrt{2+\sqrt2}\right)\\
\int_0^1\frac{\ln(1-x^{18})}{1+x^2}dx 
=& -9G+ \frac{11\pi}4\ln2 +\pi \ln(1+\sqrt3)\\
&\ +\pi\ln\left(\cos\frac\pi{36} \cos\frac{5\pi}{36} \cos\frac{7\pi}{36} \right)\\
\end{align}
A: Too long for a comment.
It's equivalent to evaluate a certain digamma series, perhaps a more attainable goal:

$$\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\psi\left(\frac{2m+1}{n}\right)$$


$\newcommand{\d}{\,\mathrm{d}}$We have, for $s>0$: $$\begin{align}J_n&:=\frac{1}{n}\int_0^1\frac{(1-t)^{s-1}}{1+t^{2/n}}t^{1/n-1}\d t\\&=\sum_{m=0}^\infty\frac{(-1)^m}{n}\int_0^1(1-t)^{s-1}t^{\frac{2m+1}{n}-1}\d t\\&=\sum_{m=0}^\infty\frac{(-1)^m}{n}\frac{\Gamma(s)\Gamma\left(\frac{2m+1}{n}\right)}{\Gamma\left(\frac{2m+1}{n}+s\right)}\end{align}$$Differentiate this w.r.t $s$: $$\sum_{m=0}^\infty\frac{(-1)^m}{n}\frac{\Gamma(s)\Gamma\left(\frac{2m+1}{n}\right)}{\Gamma\left(\frac{2m+1}{n}+s\right)}\left(\psi(s)-\psi\left(\frac{2m+1}{n}+s\right)\right)$$Evaluate at $s=1$: $$\sum_{m=0}^\infty\frac{(-1)^{m-1}}{2m+1}\left(\frac{n}{2m+1}+\psi\left(\frac{2m+1}{n}\right)+\gamma\right)=-nG-\frac{\pi\gamma}{4}-\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\psi\left(\frac{2m+1}{n}\right)$$
A: 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
$$\boxed{\int_0^1 \frac{\ln \left(1-x^{12}\right)}{1+x^2} d x= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G  + \frac{\pi}{2} \ln 6-3G= \frac{\pi}{4} \ln (72(7+4 \sqrt{3}))-6 G }$$
A: As @FShrike’s solution, I want to express the integral in terms of an infinite series of Diagamma functions.
Using $\ln \left(1-x^n\right)=-\sum_{k=1}^{\infty} \frac{x^{n k}}{k}$ for $|x|<1$, we have
$$
\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x=-\sum_{k=1}^{\infty} \frac{1}{k} \underbrace{\int_0^1 \frac{x^{n k}}{1+x^2} d x}_{I_k}
$$
$$
\begin{aligned}
I_k &=\int_0^1 \frac{y^{\frac{n k}{2}}}{1+y} \cdot \frac{1}{2 \sqrt{y}} d y \quad \textrm{, where }y=x^2\\
&=\frac{1}{2} \int_0^1 \frac{y \frac{n k-1}{2}}{1+y} d y
\end{aligned}
$$
As $\int_0^1 \frac{x^a}{1+x} d x=\frac{1}{2}\left[\psi\left(\frac{a}{2}+1\right)-\psi\left(\frac{a}{2}+\frac{1}{2}\right)\right]$, we can conclude that
$$
\boxed{
\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x =\frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k}\left[\psi\left(\frac{n k+1}{4}\right)-\psi\left(\frac{n k+3}{4}\right)\right]}
$$
