= Solution
For a <rational function>, the stipulated decay implies $R(z)=O(|z|^{-2})$ at infinity, unless it vanishes identically. Thus its real-line integral converges absolutely. Close the segment $[-T,T]$ with an upper semicircle, choosing $T$ beyond all finite poles. The arc contribution has modulus at most $\pi T\sup_{|z|=T}|R(z)|=O(T^{-1})$, tending to zero. The <residue theorem> therefore gives
$$
\boxed{\int_{-\infty}^{\infty}R(x)\,dx=2\pi i\sum_{\operatorname{Im}z_j>0}\operatorname{Res}_{z=z_j}R(z).}
$$
There are no real poles requiring indentations.
Apply this to $R(z)=1/(1+z^{2n})$. Its upper-half-plane poles are $\zeta_k=e^{i(2k+1)\pi/(2n)}$, $k=0,\ldots,n-1$, all simple. Since $\zeta_k^{2n}=-1$, their residues are
$$
\operatorname{Res}_{\zeta_k}R=\frac1{2n\zeta_k^{2n-1}}=-\frac{\zeta_k}{2n}.
$$
For $\theta=\pi/(2n)$, the finite <geometric series> gives
$$
\sum_{k=0}^{n-1}\zeta_k=e^{i\theta}\frac{1-e^{i\pi}}{1-e^{2i\theta}}=\frac{i}{\sin\theta}.
$$
The full real-line integral is consequently $\pi/[n\sin(\pi/(2n))]$. Its integrand is even, so \b[the requested half-line integral is]
$$
\boxed{\int_0^\infty\frac{dx}{1+x^{2n}}=\frac{\pi}{2n}\csc\frac{\pi}{2n}.}
$$
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