Evaluate $\int_1^\infty \frac {dx}{x^3+1}$

Since you want to evaluate this integral using contour integration method, although, I am not the expert of this method, I'll try to help you. Let $y=x-1$, then $$I=\int_1^\infty\frac{\mathrm dx}{x^3+1}=\int_0^\infty\frac{\mathrm dy}{(y+2)(y^2+y+1)}$$ Now, we can consider the contour integral \begin{align} \int_\Gamma\underbrace{\frac{\ln z}{(z+2)(z^2+z+1)}}_{f(z)}\frac{\mathrm dz}{2\pi i} =&\int^\infty_0\frac{\ln{x}-(\ln{x}+2\pi i)}{(x+2)(x^2+x+1)}\frac{\mathrm dx}{2\pi i}\\ =&-I\\ =&\color{#E2062C}{{\rm \ Res}\left(f(z),-2\right)}+\color{#00A000}{{\rm Res}\left(f(z),e^{2\pi i/3}\right)}+\color{#21ABCD}{{\rm Res}\left(f(z),e^{\color{red}{4\pi} i\color{red}{/3}}\right)}\\ =&\color{#E2062C}{\frac{1}{3}\ln{2}+\frac{\pi i}{3}}+\color{#00A000}{\frac{\pi}{3\sqrt{3}}-\frac{\pi i}{9}}\color{#21ABCD}{-\frac{2\pi}{3\sqrt{3}}-\frac{2\pi i}{9}}\\ =&-\frac{\pi}{3\sqrt{3}}+\frac{1}{3}\ln{2} \end{align} where $\Gamma$ is a keyhole contour. You may also refer to the following sources $[1]$, $[2]$, $[3]$, and $[4]$ for the rest calculation. I hope this helps.

Let's use the branch cut $\lim\limits_{R\to\infty}[1,R]$.

Now consider the integral $$ \int_\gamma\frac{\log(z-1)\,\mathrm{d}z}{1+z^3} $$ along the contour $$ \gamma=1+[\epsilon,R]e^{i\epsilon}\cup1+Re^{i[\epsilon,2\pi-\epsilon]}\cup1+[R,\epsilon]e^{-i\epsilon}\cup1+\epsilon e^{i[2\pi-\epsilon,\epsilon]} $$ As $\epsilon\to0$ and $R\to\infty$, the difference between the integral along the upper and lower lines is that $\log(z-1)$ is $2\pi i$ greater on the lower contour. Furthermore, the integral along the circular curves goes to $0$. This means that $$ \int_\gamma\frac{\log(z-1)\,\mathrm{d}z}{1+z^3}=-2\pi i\int_1^\infty\frac{\mathrm{d}x}{1+x^3} $$ $\gamma$ circles the three singularities of $\frac1{1+z^3}$ counterclockwise. Thus, the integral on the left is $2\pi i$ times the sum of the residues of $\frac{\log(z-1)}{1+z^3}$ $$ 2\pi i\left[\vphantom{\frac{\frac\pi3}3}\right.\overbrace{\frac{\frac{\pi i}3(-1-\sqrt3i)}3}^{z=e^{i\pi/3}}+\overbrace{\frac{\vphantom{\frac{\pi i}3}\log(2)+\pi i}3}^{z=-1}+\overbrace{\frac{\frac{2\pi i}3(-1+\sqrt3i)}3}^{z=e^{-i\pi/3}}\left.\vphantom{\frac{\frac\pi3}3}\right]=2\pi i\left[\frac{\log(2)}3-\frac\pi{3\sqrt3}\right] $$ Therefore, $$ \int_1^\infty\frac{\mathrm{d}x}{1+x^3}=\frac\pi{3\sqrt3}-\frac{\log(2)}3 $$