<s>[INST] Finding the definite integral of $\frac{\log x}{1+ x^3}$ The problem is apparently simple:
Find $\displaystyle{\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x }$.
However, I have not been able to get around it. The usual "big circle" strategy is useless since we don't have parity. With substitution I was able to prove that, choosing $t = 1/x$, we have
$$
\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_1^\infty \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t
$$
And, analogously
$$
\int_1^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_0^1 \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t
$$
However, this doesn't seem to be helpful. I could now write my integral as \begin{align}
\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x= &
\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x + \int_1^\infty \!\! \frac{\log x}{1+ x^3} \,\mathrm{d} x \\ 
= &\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x - \int_0^1 \!\!\frac{x \log x}{1+ x^3} \,\mathrm{d} x 
\\ 
= &\int_0^1 \!\!\frac{1-x}{1+ x^3} \log x \,\mathrm{d} x\\
\end{align}
This is a proper definite integral and I hoped it could be useful, e.g. to use Integration by series, but I didn't manage to do it.
 [/INST] Continue with
\begin{align}
\int_0^\infty&\frac{\log x}{1+x^3}{d}x
= 
\int_0^1\frac{(1-x)\log x}{1+x^3}{d}x \\
&=\int_0^1\frac{\log x}{1+x}
 - \underset{x^3\to x}{\frac{x^2\log x}{1+x^3}}\ {dx}
= \frac89\int_0^1\frac{\log x}{1+x}{d}x
=-\frac{2\pi^2}{27}
\end{align}
$\int_0^1\frac{\log x}{1+x}{d}x=-\frac{\pi^2}{12}$
</s>