Show that $$\int_0^\infty\frac{1}{1+x^n}\,\mathrm dx = \frac{\pi/n}{\sin(\pi/n)}$$ for $\mathbb{N}\ni n\geq 2$.
Let $S=\{r\mathrm e^{\mathrm i\varphi}\in\mathbb{C} \mid 0\leq r\leq R,0\leq \varphi\leq (2\pi)/n\}$ and we will integrate along $\partial S$ for $R\to\infty$. With $\gamma(t)=R\mathrm e^{\mathrm it}, t\in[0,(2\pi)/n]$ and $\delta=[R\mathrm e^{\mathrm i \frac{2\pi}{n}}, 0]$ we can write this integral as
$$ \oint_{\partial S}\frac{1}{1+z^n}\,\mathrm dz = \underbrace{\int_0^R \frac{1}{1+z^n}\,\mathrm dz}_{=: I_R} + \int_\gamma\frac{1}{1+z^n}\,\mathrm dz + \int_{\delta} \frac{1}{1+z^n}\,\mathrm dz. $$
Fortunately the integral along the outer arc $\gamma$ vanishes since
$$ \left|\int_\gamma\frac{1}{1+z^n}\,\mathrm dz\right| \leq \int_0^{(2\pi)/n}\left|\frac{R\mathrm i\mathrm e^{\mathrm it}}{R^n\mathrm e^{n\mathrm it}+1}\right|\,\mathrm dt \overset{(*)}{\leq} \frac{R}{(R-1)^n}\frac{2\pi}{n} \overset{R\to\infty}{\longrightarrow} 0. $$
Furthermore we have
$$ \int_{\delta} \frac{1}{1+z^n}\,\mathrm dz = - \int_0^R \frac{\mathrm e^{\mathrm i \frac{2\pi}{n}}}{(t\mathrm e^{\mathrm i \frac{2\pi}{n}})^n+1}\,\mathrm dt = - \int_0^R \frac{\mathrm e^{\mathrm i \frac{2\pi}{n}}}{t^n+1}\,\mathrm dt = - \mathrm e^{\mathrm i \frac{2\pi}{n}}I_R. $$
The singularities of the integrand are obtained via $z^n+1=0\Leftrightarrow z_k=\mathrm e^{\mathrm i \frac{\pi}{n}}\mathrm e^{\mathrm i k\frac{2\pi}{n}}$ for all $k=0,\ldots,n-1$. Since only $z_0$ lies in $S$ this is the only singularity we have to consider for the residue theorem. It follows that
$$ \oint_{\partial S}\frac{1}{1+z^n}\,\mathrm dz = 2\pi\mathrm i\operatorname{Res}_{z_0}\left(\frac{1}{1+z^n}\right) = 2\pi\mathrm i\frac{1}{nz_0^{n-1}} = 2\pi\mathrm i\frac{1}{n\mathrm e^{\mathrm i \frac{(n-1)\pi}{n}}} = -\frac{2}{n}\pi\mathrm i\mathrm e^{\mathrm i \frac{\pi}{n}}. $$
We are interested in the value $I=\lim_{R\to\infty} I_R$ hence we can write
$$ -\frac{2}{n}\pi\mathrm i\mathrm e^{\mathrm i \frac{\pi}{n}} = I - \mathrm e^{\mathrm i \frac{2\pi}{n}}I. $$
Now we can solve for $I$ which yields
$$ I = \frac{-\frac{2}{n}\pi\mathrm i\mathrm e^{\mathrm i \frac{\pi}{n}}}{1 - \mathrm e^{\mathrm i \frac{2\pi}{n}}} = \frac{\pi}{n}2\mathrm i\frac{- \mathrm e^{\mathrm i \frac{\pi}{n}}}{1- \mathrm e^{\mathrm i \frac{2\pi}{n}}} = \frac{\pi}{n}\frac{2\mathrm i}{\mathrm e^{\mathrm i \frac{\pi}{n}} - \mathrm e^{-\mathrm i \frac{\pi}{n}}} = \frac{\pi/n}{\sin(\pi/n)}. $$
I am curious whether this suffices and/or you have other suggestions to further improve my solution. Furthermore I am indecisive about my inequality at $(*)$ regarding the vanishing integral along $\gamma$. This inequality for example does not hold for $R=1/2$ and $n=3$ - is it reasonable to assume that $R\geq 1$ since we are analyzing $R\to\infty$ anyways? If so I would argue that $R\geq 1\implies R^n-1>0$ and therefore $R^n-1\geq (R-1)^n$. Are there alternative justifications for this estimation?
Without copmplex analysis: substitute
$$u:=\frac{x^n}{1+x^n}\iff\implies du=\frac{nx^{n-1}(1+x^n)-nx^{2n-1}}{(1+x^n)^2}dx=\frac{n\frac{u^{1-\frac1n}}{(1-u)^{1-\frac1n}}\frac1{1-u}-n\frac{u^{2-\frac1n}}{(1-u)^{2-\frac1n}}}{\frac1{(1-u)^2}}dx$$
$$=n\left(u^{1-\frac1n}(1-u)^{\frac1n}-u^{2-\frac1n}(1-u)^{\frac1n}\right)dx=n(1-u)^{1+\frac1n}u^{1-\frac1n}dx\implies$$
$$\int_0^\infty\frac{dx}{1+x^n}=\frac1n\int_0^1(1-u)^{\frac1n}u^{\frac1n-1}du=\frac1n\overbrace{B\left(\frac1n\,,\,\,1-\frac1n\right)}^{\text{def. of Beta Funct.}}=$$
$$=\frac1n\frac{\Gamma\left(\frac1n\right)\Gamma\left(1-\frac1n\right)}{\Gamma(1)}\stackrel{\text{Euler's Reflection Form.}}=\frac1n\frac\pi{\sin\frac\pi n}=\cfrac{\cfrac\pi n}{\sin\cfrac\pi n}$$