Here are two approaches.
Beta Function and Euler's Reflection Formula
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^n}
&=\frac1n\int_0^\infty\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag{1a}\\
&=\frac1n\,\Gamma\!\left(\frac1n\right)\Gamma\!\left(1-\frac1n\right)\tag{1b}\\[3pt]
&=\frac\pi n\csc\left(\frac\pi n\right)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $x\mapsto x^{1/n}$
$\text{(1b)}$: Beta Integral
$\text{(1c)}$: Euler's Reflection Formula
Applying Elementary (Completely Real) Tools
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{1+x^n}
&=\frac1n\int_0^\infty\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag{2a}\\
&=\frac1n\int_0^1\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x+\frac1n\int_0^1\frac{x^{-\frac1n}}{1+x}\,\mathrm{d}x\tag{2b}\\
&=\frac1n\int_0^1\sum_{k=0}^\infty(-1)^kx^{k+\frac1n-1}\,\mathrm{d}x+\frac1n\int_0^1\sum_{k=0}^\infty(-1)^kx^{k-\frac1n}\,\mathrm{d}x\tag{2c}\\
&=\frac1n\sum_{k=0}^\infty\frac{(-1)^k}{k+\frac1n}+\frac1n\sum_{k=0}^\infty\frac{(-1)^k}{k-\frac1n+1}\tag{2d}\\
&=\frac1n\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\frac1n}\tag{2e}\\[3pt]
&=\frac\pi{n}\csc\left(\frac\pi{n}\right)\tag{2f}
\end{align}
$$
Explanation:
$\text{(2a)}$: substitute $x\mapsto x^{1/n}$
$\text{(2b)}$: break the domain of integration into $[0,1]$ and $[1,\infty)$
$\phantom{\text{(2b):}}$ substitute $x\mapsto1/x$ on $[1,\infty)$
$\text{(2c)}$: Taylor series for $\frac1{1+x}$
$\text{(2d)}$: integrate
$\text{(2e)}$: substitute $k\mapsto-k-1$ in the right-hand sum
$\text{(2f)}$: apply $(8)$ from this answer