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.
Best Answer
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 $\int_0^{\pi/4}\ln (2\cos 2t)dt = \int_0^{\pi/4}\ln (2\sin 2t) d t=0$.