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.
 A: Substitute $x=\tan t$. Then
$$1-x^4=4x^2\frac{1-x^2}{1+x^2}\left(\frac{1+x^2}{2x}\right)^2
=4\tan^2t\ \frac{\cos 2t}{\sin^2 2t}
$$
and
\begin{aligned}
&\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x \\
=& \int_0^{\pi/4}\ln8+2\ln \tan t +\ln (2\cos 2t)
-2\ln (2\sin 2t) \ d t\\
=&\ \frac{3\pi}4\ln2 - 2G
\end{aligned}
where $\int_0^{\pi/4} \ln \tan t\ dt=-G$ and
$\int_0^{\pi/4}\ln (2\cos 2t)dt
= \int_0^{\pi/4}\ln (2\sin 2t)  d t=0$.
A: 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
$$\int_{0}^{1}\frac{\ln\left(1-x^{4}\right)}{1+x^{2}}dx = \frac{3\pi}{4}\ln\left(2\right)-2G.$$
A: Letting $x=\tan \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4} } \ln \left[\left(1+\tan ^2 \theta\right)\left(1-\tan ^2 \theta\right)\right] d \theta \\
&=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta+\ln \left(\frac{\cos^2 \theta-\sin ^2 \theta}{\cos ^2 \theta}\right) d \theta \\
&=-4 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta+\int_0^{\frac{\pi}{4}} \ln (\cos 2 \theta) d \theta \\
&=-4\left(-\frac{\pi}{4} \ln 2+\frac{G}{2}\right)-\frac{\pi}{4} \ln 2 \\
&=\frac{3 \pi}{4} \ln 2-2G
\end{aligned}
$$
where the first integral refers to the post.
