= Solution
The <residue theorem> states that the <contour integral> of a <meromorphic function> around a positively oriented simple closed contour equals $2\pi i$ times the sum of its enclosed <residues>, provided there are no singularities on the contour. The roots of $1+z^n$ are $z_k=e^{(2k+1)i\pi/n}$, $k=0,\ldots,n-1$. They are simple <poles>, with <residues> $1/(nz_k^{n-1})=-z_k/n$.
Use the sector contour of angle $2\pi/n$, running outward along the positive axis and inward along the rotated ray. It encloses only $z_0=e^{i\pi/n}$. The large-arc contribution is $O(R^{1-n})$, hence vanishes. With $J=\int_0^\infty dx/(1+x^n)$, the two ray contributions give
$$
(1-e^{2\pi i/n})J=-\frac{2\pi i}{n}e^{i\pi/n}.
$$
Since $1-e^{2i\alpha}=-2ie^{i\alpha}\sin\alpha$,
$$
\boxed{J=\frac\pi{n\sin(\pi/n)}.}
$$
For the second integral, use $z^{n/2}/(1+z^n)$ on a branch continuous across the sector away from zero, with a small inner arc of radius $\epsilon$. Its large arc is $O(R^{1-n/2})$ and its small arc is $O(\epsilon^{1+n/2})$, both vanishing for $n\geq3$. On the rotated ray, $z^{n/2}$ changes by $e^{i\pi}=-1$, so the two radial contributions add rather than subtract. The enclosed <residue> is $z_0^{1-n/2}/n=-ie^{i\pi/n}/n$. Consequently
$$
(1+e^{2\pi i/n})I=\frac{2\pi}{n}e^{i\pi/n},\qquad\boxed{I=\frac\pi{n\cos(\pi/n)}}.
$$
Both answers are instances of the <Mellin integral of a reciprocal binomial>. The direct contour argument also tracks the branch needed when $n$ is odd.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2-sector-contour.png]
{title=A sector encloses exactly one root of the denominator}
{description=An original illustrative contour for $n=5$, with outward and inward rays, an outer arc, and the inner indentation needed for a fractional power. Only the root at angle $\pi/n$ is enclosed.}
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