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?

Letting $x\mapsto \frac{1}{x}$ simplify the integral as
$$
I=\int_0^{\infty} \ln \left(\frac{1+x^3}{x^3}\right) \frac{x}{1+x^3} d x =\int_0^{\infty} \frac{\ln \left(x^3+1\right)}{x^3+1} d x
$$
Next consider another integral
$$
I(a)=\int_0^{\infty}\left(x^3+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^3+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{3} \int_0^1 y^{-a-\frac{4}{3}}(1-y)^{-\frac{2}{3}} d y \\
&=\frac{1}{3} B\left(-a-\frac{1}{3}, \frac{1}{3}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{3} B\left(-a-\frac{1}{3}, \frac{1}{3}\right)\left(\psi\left(-a-\frac{1}{3}\right)-\psi(-a)\right)
$$
Then putting $a=1$ gives our integral$$
\begin{aligned}
I&=I^{\prime}(-1) \\&=\frac{1}{3} B\left(\frac{2}{3}, \frac{1}{3}\right)\left(\psi\left(\frac{2}{3}\right)-\psi(1)\right) \\
&=\frac{1}{3} \cdot \frac{2 \pi}{\sqrt{3}}\left(\gamma-\frac{\pi}{2 \sqrt{3}}+\frac{3 \ln 3}{2}-\gamma\right) \\
&=\frac{\pi}{\sqrt{3}} \ln 3-\frac{\pi^2}{9}
\end{aligned}
$$
