= Solution
The <improper integral> converges: its integrand is bounded near zero and at most $x^{-n}$ for $x\geq1$, which is integrable because $n\geq2$.
Set $\theta=\pi/n$, $\omega=e^{2i\theta}$ and $\zeta=e^{i\theta}$. Apply the <residue theorem> to $F(z)=1/(1+z^n)$ on the positively oriented sector contour consisting of the segment $0\to R$, the arc $Re^{it}$ for $0\leq t\leq2\theta$, and the segment $R\omega\to0$, with $R>1$. No <pole> lies on the boundary. Exactly one root of $z^n=-1$ is inside this sector, namely $\zeta$; it is a <simple pole>. Part (a) gives
$$
\operatorname{Res}_{\zeta}F=\frac1{n\zeta^{n-1}}=-\frac\zeta n.
$$
On the returning ray, put $z=\omega x$ with $x$ decreasing from $R$ to zero. Since $\omega^n=1$, the two ray contributions are
$$
\int_0^R\frac{dx}{1+x^n}+\int_R^0\frac{\omega\,dx}{1+x^n}=(1-\omega)I_R,
\qquad I_R=\int_0^R\frac{dx}{1+x^n}.
$$
On the arc, $|1+z^n|\geq R^n-1$, so the <estimation lemma> bounds its <contour integral> by
$$
\frac{2\pi R}{n(R^n-1)}\longrightarrow0.
$$
The <residue theorem> and the convergence already established now give
$$
(1-e^{2i\theta})I=-\frac{2\pi i}{n}e^{i\theta}.
$$
Using $1-e^{2i\theta}=-2ie^{i\theta}\sin\theta$, we conclude
$$
\boxed{\int_0^\infty\frac{dx}{1+x^n}=\frac\pi{n\sin(\pi/n)}\qquad(n\geq2)}.
$$
The sector also works for $n=2$, when it is the upper semicircle. Its angle is chosen so that $z^n$ has the same value on the two rays.
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