Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x^2} d x}_K+ \underbrace{\int_0^1 \frac{\ln (1-x)}{1+x^2} d x}_L
\end{aligned}

Denotes the Catalan’s constant by $G$.
By my post, $$J=\frac{\pi}{2}\ln2-G$$
Dealing with the last $2$ integrals, we use a powerful substitution $x=\frac{1-t}{1+t} ,$ then $dx=-\frac{2dt}{(1+t)^2}.$

$$
\begin{aligned}
K&=\int_1^0 \frac{\ln 2-\ln (1+t)}{\frac{2+2 t^2}{(1+t)^2}} \frac{-2 d t}{(1+t)^2} \\
&=\ln 2 \int_0^1 \frac{d t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2}
\end{aligned}
$$
Hence $$K=\frac{\pi}{8} \ln 2 $$

$$
\begin{aligned}
L=& \int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\
&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t . \\
&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G
\end{aligned}
$$

Combining them to get
$$
\begin{aligned}
I &=\left(\frac{\pi}{2} \ln 2-G\right) +\frac{\pi}{8} \ln 2  +\left(\frac{\pi}{8} \ln 2-G\right)\\
&=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
I do want to know if it can be solved by any other elegant methods. Your comments and methods are highly appreciated.

This approach is somewhat similar to yours. It is more of a comment than an answer, but it is too long to fit in the comments section.
Let the integral in question be $I$. Letting $x \to \dfrac{1-x}{1+x}$ at the beginning yields
$$I = 2\int_{0}^{1}\frac{\ln\left(1-\left(\frac{1-x}{1+x}\right)^{4}\right)}{1+\left(\frac{1-x}{1+x}\right)^{2}}\left(\frac{1}{\left(1+x\right)^{2}}\right)dx$$
which simplifies down to
$$3\ln\left(2\right)\int_{0}^{1}\frac{1}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(x\right)}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(1+x^{2}\right)}{1+x^{2}}dx-4\int_{0}^{1}\frac{\ln\left(1+x\right)}{1+x^{2}}dx.$$
Evaluating the first integral trivially and using the assumptions from your query, we get
$$I = \frac{3}{4}\pi\ln\left(2\right) - G + \frac{\pi}{2}\ln\left(2\right)-G - 4\left(\frac{\pi}{8}\ln\left(2\right)\right)$$
Therefore, the integral $I$ is
$$\int_{0}^{1}\frac{\ln\left(1-x^{4}\right)}{1+x^{2}}dx = \frac{3\pi}{4}\ln\left(2\right)-2G.$$
