Resembling Resemblence-3

Define

$\color{#D61F06}{I(a)=\displaystyle\int_0^\infty \log \dfrac{a+x^3}{x^3} \dfrac{x \,dx}{1+x^3}}$

Then $I(0)=0$ and

$\color{#3D99F6}{\begin{aligned} I'(a)&=&\displaystyle\int_0^\infty\dfrac{x}{(a+x^3)(1+x^3)}dx\\ &=&\dfrac13\displaystyle\int_0^\infty\dfrac{1}{x^{1/3}(a+x)(1+x)}dx\\ &=&\dfrac{1}{3(1-a)}\left(\displaystyle\int_0^\infty\dfrac{1}{x^{1/3}(a+x)}dx-\int_0^\infty\dfrac{1}{x^{1/3}(1+x)}dx\right)\\ &=&\dfrac{1}{3(1-a)}\dfrac{2\pi}{\sqrt3}\left(\frac{1}{a^{1/3}}-1\right)\\ &=&\dfrac{2\pi}{3\sqrt3}\dfrac{1}{a+a^{2/3}+a^{1/3}} \end{aligned}}$

Here we use

$\color{#20A900}{\displaystyle\int_0^\infty\dfrac{1}{x^{1/3}(a+x)}dx=\dfrac{2\pi}{\sqrt3 a^{1/3}}}$

Thus

$\color{#3D99F6}{\begin{aligned} I(1)&=&\frac{2\pi}{\sqrt3}\int_0^1 \frac{1}{a+a^{2/3}+a^{1/3}}da\\ &=&\frac{2\pi}{3\sqrt3}\int_0^1 \frac{b}{b^2+b+1}db\\ &=&-\boxed{\dfrac{\pi^2}{9}+\dfrac{\pi}{\sqrt3}\ln 3}. \end{aligned}}$