contour integration of logarithm I must compute the following integral
$$\displaystyle\int_{0}^{+\infty}\frac{\log x}{1+x^3}dx$$
Can someone suggest me the right circuit in the complex plane over which to do the integration? I tried different paths, avoiding the origin, but unsuccessfully

Let
$$I = \int_0^{\infty} \dfrac{\log(x)}{1+x^3} dx = \underbrace{\int_0^1 \dfrac{\log(x)}{1+x^3} dx}_J + \underbrace{\int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx}_K$$
$$K = \int_1^{\infty} \dfrac{\log(x)}{1+x^3} dx = \int_1^0 \dfrac{\log(1/x)}{1+1/x^3} \left(-\dfrac{dx}{x^2}\right) = - \int_0^1 \dfrac{x \log(x)}{1+x^3} dx$$
Now recall that
$$\int_0^1 x^m \log(x) dx = - \dfrac1{(m+1)^2}$$
Hence,
$$J = \int_0^1 \dfrac{\log(x)}{1+x^3} dx = \int_0^1 \sum_{k=0}^{\infty}(-x^3)^k \log(x) dx = \sum_{k=0}^{\infty}(-1)^k \int_0^1 x^{3k} \log(x) dx = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(3k+1)^2}$$
$$K = -\int_0^1 \dfrac{x\log(x)}{1+x^3} dx = \sum_{k=0}^{\infty}(-1)^{k+1} \int_0^1 x^{3k+1} \log(x) dx = \sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{(3k+2)^2}$$
Hence,
$$J = -\sum_{k=0}^{\infty} \dfrac1{(6k+1)^2} + \sum_{k=0}^{\infty} \dfrac1{(6k+4)^2} = f_4 - f_1$$
$$K = \sum_{k=0}^{\infty} \dfrac1{(6k+2)^2} - \sum_{k=0}^{\infty} \dfrac1{(6k+5)^2} = f_2 - f_5$$
where $$f_l = \sum_{k=0}^{\infty} \dfrac1{(6k+l)^2}$$
Note that $f_6 = \dfrac{\pi^2}{216}$, $f_3 = \dfrac19 \cdot \dfrac{\pi^2}8 = \dfrac{\pi^2}{72}$. Let $\zeta$ be the $6^{th}$ root of unity i.e. $\zeta = e^{\pi i /3}$.
We have
$$\text{Li}_2(\zeta) = \sum_{k=1}^{\infty} \dfrac{\zeta^k}{k^2} = \zeta f_1 + \zeta^2 f_2 - f_3 - \zeta f_4 - \zeta^2 f_5 + f_6 = - \dfrac{\pi^2}{108} - \zeta J + \zeta^2 K$$
$$\text{Li}_2(\zeta^5) = \sum_{k=1}^{\infty} \dfrac{\zeta^{5k}}{k^2} = \zeta^5 f_1 + \zeta^{10} f_2 + \zeta^{15} f_3 + \zeta^{20} f_4 + \zeta^{25} f_5 + f_6 = - \dfrac{\pi^2}{108} + \zeta^2 J - \zeta K$$
where $\text{Li}_s(x)$ is the polylogarithm function defined as
$$\text{Li}_s(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k^s}$$
Polylgarithm function satisfies a nice identity namely
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ are Bernoulli polynomials. Take $n=2$ and $x = 1/6$ to get that
$$\text{Li}_2(\zeta) + \text{Li}_2(\zeta^5) = - \dfrac{(2\pi i)^2}{2!}B_2(1/6) = - \dfrac{(2\pi i)^2}{2!} \dfrac1{36} = \dfrac{\pi^2}{18}$$
Hence,
\begin{align}
\zeta^2(J+K) - \zeta(J+K) - \dfrac{\pi^2}{54} & = \dfrac{\pi^2}{18}\\
\overbrace{(\zeta^2-\zeta)}^{-1}(J+K) & = \dfrac{\pi^2}{18} + \dfrac{\pi^2}{54}\\
-(J+K) & = \dfrac{2\pi^2}{27}
\end{align}
Hence,
$$I = J+K = -\dfrac{2 \pi^2}{27}$$
