How to evaluate this integral using contour integration( Help needed in choosing contour) This question is part of a assignment of complex analysis and I was unable to solve the following question.

Using contour integration evaluate the following : $\int_{0}^{\infty} \frac{1} {1+ x^6} dx$.

By De Moivres's theorem one can find roots of $1+x^6$ but I am unable to decided how to choose the contour. Actually, I am new to such questions which involve contour integration.
I searched MSE for quite some time and found some questions on contour integration like this: Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$ but I am not able to understand the solution of it.
Before I found it in the assignment I had a question of contour integration like this : $\int_{-\infty}^{\infty} \frac{1}{1+x^2}$  and I evaluated that sucessfully but I am unable to apply that here due to number of roots being 6 and I am having trouble in choosing contour .

So, can you please help with choosing the contour in this question and what details should be kept in mind while choosing contour?

 A: $$I=\int\limits_0^\infty\dfrac{\text dx}{1+x^6}
=\dfrac12\int\limits_{-\infty}^\infty f(x)\,\text dx,$$
where $\,f(x)=\dfrac{1}{1+x^6}\,$ is the even function.
Let us consider the closed contour $\,\Gamma=L+C,\,$ where $\,L\,$ is the interval $\,[-R,R]\,$ (purple line) and $\,C\,$ is the half-circle with the radius $\,R>1,\,$ considered by increasing of polar angle $\,\varphi\,$ from $\,0\,$ to $\,\pi\,$ (blue line).

The function $\,f(z),\;z\in\mathbb C,\,$ has six poles, which can be founded from the equation $\,z^6=-1=e^{i\pi}.\,$ Assuming a polar notation $\,z=\rho e^{i\varphi},\;\rho\in(0,\infty),\;\varphi\in[0,2\pi),\,$ easily to get
$$\rho^6=1,\quad 6\varphi =\pi+2\pi k,\quad k\in\mathbb Z,$$
with the set of solutions
$$\rho=1,\;\varphi\in\left\{\dfrac{\pi}6, \dfrac{\pi}2, \dfrac{5\pi}6, \dfrac{7\pi}6, \dfrac{3\pi}2, \dfrac{11\pi}6\right\},$$
$$z\in\left\{\dfrac{\sqrt3}2+\dfrac12i, i, -\dfrac{\sqrt3}2+\dfrac12i,
-\dfrac{\sqrt3}2-\dfrac12i, -i, \dfrac{\sqrt3}2-\dfrac12i\right\},$$
wherein the poles
$$z_0=e^{i\,\large^\pi/_6}=\dfrac{\sqrt3}2+\dfrac12i,\quad z_1=e^{i\,\large^\pi/_2}=i,\quad z_2=e^{i\,\large^{5\pi}/_6}=-\dfrac{\sqrt3}2+\dfrac12i$$
belong to the inner part of the contour  $\,\Gamma.\,$
Since the poles $\,z_0,z_1,z_2\,$ are simple and the contour $\,\Gamma\,$ does not contain additional singularities, then for arbitrary $\,R\in(1,\infty)\,$
$$J(R)=\oint\limits_\Gamma f(z)\text dz
=2\pi i\sum\limits_{j=0,1,2}\underset{z_j}{\text{Res}}\,\dfrac1{1+z^6}
=2\pi i\sum\limits_{j=0,1,2}\lim\limits_{z\to z_j}\,\dfrac1{6z^5}
=\dfrac\pi3 i\sum\limits_{j=0,1,2}\dfrac {z_j}{z_j^6},$$
$$J(R)=-\dfrac\pi3 i\sum\limits_{j=0,1,2}z_j
=-\dfrac\pi3 i\left(\dfrac{\sqrt3}2+\dfrac12i+i-\dfrac{\sqrt3}2+\dfrac12i\right)=\dfrac23\pi.$$
On the other hand,
$$J(R)=\int\limits_{-R}^R f(x)\,\text dx + iR\int\limits_0^\pi f(Re^{i\varphi})e^{i\varphi}\,\text d\varphi,$$
wherein
$$\lim\limits_{R\to\infty}\left|iR\int\limits_0^\pi f(Re^{i\varphi})e^{i\varphi}\,\text d\varphi\right|
\leq \lim\limits_{R\to\infty}\pi\dfrac R{R^6-1}=0,$$
$$J(\infty)=2I,$$
$$\color{green}{\mathbf{I=\dfrac\pi3.}}$$
