<s>[INST] Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}
It checks out numerically, for $n=1,\dots,7$ that
\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}
so \eqref{0} seems to follow the pattern
\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}
for some function $f$.
Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.

Edit:
Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}
and
\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}
with the same $f$.

Edit
Thanks to the great answer by @Quanto,
the function $f$ can be defined as
\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}
$\endgroup$
 [/INST] Not a complete answer, but an elaboration of what I said in the comments.
To evaluate $$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$$
Write it as $$\int_1^\infty\frac{\ln\left(x^{-2n}+1\right)+\ln\left(x^{2n}\right)}{1+x^2} dx = \int_1^\infty\frac{\ln\left(x^{-2n}+1\right)}{1+x^2} dx+2n\int_1^\infty\frac{\ln\left(x\right)}{1+x^2} dx$$
The second integral is a standard integral for Catalan's constant. To solve the first, make the substitution $x \to \frac{1}{x}$ to get $$\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx + 2nG$$
So if you know the value of any of the integrals with bounds $(0,\infty)$ or $(1,\infty)$ or $(0,1)$, you could find the other two.

The series for $\ln(1+x)$ converges for $|x|<1$, so expand it to get $$\int_0^1\frac{\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}x^{2nk}}{1+x^2}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\frac{x^{2nk}}{1+x^{2}}dx$$
Then you can expand $\frac{1}{1+x^2}$ to get $$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}x^{2nk}\sum_{m=0}^{\infty}\left(-1\right)^{m}x^{2m}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\int_{0}^{1}x^{2nk}x^{2m}dx$$
Finally, evaluate the inner integral to get
$$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\frac{1}{2nk+2m+1}$$
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