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?

Let us make the change of variables 
$$v=\frac{x^3}{1+x^3}\iff x=\left(\frac{v}{1-v}\right)^{1/3}$$
This transforms the integral $I$ to the following form
$$
I=-\frac{1}{3}\int_0^1\log(v)\,v^{-1/3}(1-v)^{-2/3}dv
$$
Now, If 
$$f(\alpha):=B(\alpha,\frac{1}{3})=\int_0^1v^{\alpha-1}(1-v)^{\frac{1}{3}-1}dv=\frac{\Gamma(\alpha)\Gamma(\frac{1}{3})}{\Gamma(\alpha+\frac{1}{3})}$$
then $I=-\frac{1}{3}f'(\frac{2}{3})$.
But, since $\Gamma(1)=1$, and $\Gamma'(1)=-\gamma$, we have
$$\eqalign{
f'\left(\frac{2}{3}\right)&=\Gamma'\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)-\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\Gamma'(1)\cr
&=\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr
&\buildrel{\rm(1)}\over{=}\frac{\pi}{\sin(\pi/3)}\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr
&\buildrel{\rm(2)}\over{=}\frac{2\pi}{\sqrt{3}}\left(
\frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right)
}
$$
Where, for $(1)$ we used the Euler's reflection formula, and for $(2)$ we used Gauss' theorem for the digamma function. Combining our results we get
$$
I=-\frac{1}{3}\frac{2\pi}{\sqrt{3}}\left(
\frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right)=\frac{\pi}{\sqrt{3}}\log 3-\frac{\pi^2}{9}.
$$
which is the desired result.$\qquad\square$
