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 $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
Best Answer
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}