Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
  We can possibly try
$$
I=\int_0^\infty \frac{\log(1+x^3) x}{1+x^3} \,dx-\int_0^\infty \frac{\log(x^3) x}{1+x^3}\,dx.
$$ These integrals would be much easier had the bounds been from $0 $\ to $\infty$, however they are not.  Perhaps partial integration will work but I didn't find the way if we try
$$
dv=\frac{x}{1+x^3}, \quad u= \log(1+x^3)
$$
but I ran into a divergent integral. Thanks how can we prove I?

Define
$$
I(a)=\int_0^\infty \log \frac{a+x^3}{x^3} \frac{x \,dx}{1+x^3}.$$
Then $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^\infty\frac{x}{(a+x^3)(1+x^3)}dx\\
&=&\frac13\int_0^\infty\frac{1}{x^{1/3}(a+x)(1+x)}dx\\
&=&\frac{1}{3(1-a)}\left(\int_0^\infty\frac{1}{x^{1/3}(a+x)}dx-\int_0^\infty\frac{1}{x^{1/3}(1+x)}dx\right)\\
&=&\frac{1}{3(1-a)}\frac{2\pi}{\sqrt3}\left(\frac{1}{a^{1/3}}-1\right)\\
&=&\frac{2\pi}{3\sqrt3}\frac{1}{a+a^{2/3}+a^{1/3}}
\end{eqnarray}
Here we use
$$ \int_0^\infty\frac{1}{x^{1/3}(a+x)}dx=\frac{2\pi}{\sqrt3 a^{1/3}}. $$
Thus
\begin{eqnarray}
I(1)&=&\frac{2\pi}{\sqrt3}\int_0^1 \frac{1}{a+a^{2/3}+a^{1/3}}da\\
&=&\frac{2\pi}{3\sqrt3}\int_0^1 \frac{b}{b^2+b+1}db\\
&=&-\frac{\pi^2}{9}+\frac{\pi}{\sqrt3}\ln 3.
\end{eqnarray}
