= Solution
Let $\alpha=\pi(m+1)/n$, which lies strictly between zero and $\pi$. On the lower ray $z=x$, the integral is $I_R=\int_0^R x^m/(1+x^n)dx$. On the upper ray $z=xe^{2\pi i/n}$, the positive boundary orientation runs from $R$ down to zero. Since $z^n=x^n$, that contribution is $-e^{2i\alpha}I_R$. The circular contribution vanishes by part (b); hence
$$
(1-e^{2i\alpha})I=-\frac{2\pi i}{n}e^{i\alpha}.
$$
Use $1-e^{2i\alpha}=-2ie^{i\alpha}\sin\alpha$ to obtain
$$
\boxed{\int_0^\infty\frac{x^m}{1+x^n}\,dx
=\frac\pi{n\sin[\pi(m+1)/n]}}.
$$
The result is positive as expected. Convergence near zero follows from $m\ge0$, so both endpoint limits used in the <contour> argument are justified.
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