Integrating $\int_0^{\infty} \frac{dx}{1+x^3}$ using residues.

Since $(e^{2\pi i/3}z)^3 = z^3$, a suitable contour is a third of a circle, with the rays on the positive real axis and $[0,\infty)\cdot e^{2\pi i/3}$. That gives you

$$2\pi i \operatorname{Res}\left(\frac{1}{1+z^3}; e^{\pi i/3}\right) = \int_0^\infty \frac{dx}{1+x^3} - e^{2\pi i/3}\int_0^\infty \frac{dx}{1+(e^{2\pi i/3}x)^3}.$$

From then on, it's a simple manipulation.

I know this is long overdue, but:

We're looking to take advantage of the behavior of the complex logarithm (at the branch cut). Consider the following contour integral: $$\oint_C \frac{\ln z}{1+x^3}dz=2\pi i\sum\text{Res}f(z)=\int_0^\infty\frac{\ln z}{1+z^3}dz+\int_\infty^0 \frac{\ln z+2\pi i}{1+z^3}dz=-2\pi i\int_0^\infty\frac{1}{1+z^3}dz$$

(where $C$ is the well-known keyhole contour). The residues are found using the following expression:

$$\frac{h(z_0)}{g'(z_0)}=\frac{\ln z_0}{(1+z_0^3)'}=\frac{\ln z_0}{3z_0^2}=\frac{z_0\ln z_0}{3z_0^3}=-\frac{z_0\ln z_0}{3} \tag{$\forall z_0, z_0^3=-1$}$$ The sum of residues is then: $$\sum\text{Res}f(z)=-\ln(e^{i\pi/3})\frac{e^{i\pi/3}}3-\ln(e^{i\pi})\frac{e^{i\pi}}3-\ln(e^{5i\pi/3})\frac{e^{5i\pi/3}}3$$ $$=-\frac{i\pi}{9}\left(\frac 12 + i\frac{\sqrt 3}2\right)+\frac{i\pi}3-\frac{5\pi i}{9}\left(\frac 12 - i\frac{\sqrt 3}2\right)$$ $$=-{\frac{i\pi}{18}}+\frac{\pi\sqrt 3}{18}+\frac{6i\pi}{18}-\frac{5i\pi}{18}-\frac{5\pi\sqrt 3}{18}$$ $$=-\frac{2\sqrt 3}{9}\pi$$

$$2\pi i\sum\text{Res}f(z)=-2\pi i\int_0^\infty\frac{1}{1+z^3}dz$$ $$\boxed{\color{blue}{\int_0^\infty\frac{1}{1+z^3}dz=-\sum\text{Res}f(z)=\frac{2\sqrt 3}{9}\pi=\frac{2}{3\sqrt 3}\pi}}$$

not $$I=\int_{0}^{+\infty}\dfrac{1}{1+x^3}dx=\dfrac{1}{3}B\left(\dfrac{1}{3},1-\dfrac{1}{3}\right)=\dfrac{1}{3}\dfrac{\Gamma{(\dfrac{1}{3})}\Gamma{(\dfrac{2}{3})}}{\Gamma{(1)}}=\dfrac{2\pi}{3\sqrt{3}}$$