= Solution
{c}
Extend the integrand by zero to all of $(0,\infty)$:
$$
h_n(x)=\mathbf1_{(0,2n\pi)}(x)\frac{(n+x)\sin(x/n)}{x(1+x^2)}.
$$
For each fixed $x>0$, eventually the <indicator function> is one and $\sin(x/n)/(x/n)\to1$. Hence $h_n(x)\to(1+x^2)^{-1}$ <pointwise>. On its support, $x/n\leq2\pi$, so $|\sin t|\leq t$ for $t\geq0$ gives
$$
|h_n(x)|\leq\frac{1+x/n}{1+x^2}\leq\frac{1+2\pi}{1+x^2}.
$$
The final bound is <Lebesgue integrable> on $(0,\infty)$ and independent of $n$. The <dominated convergence theorem> therefore gives
$$
\boxed{\lim_{n\to\infty}\int_0^{2n\pi}\frac{(n+x)\sin(x/n)}{x(1+x^2)}\,dx
=\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2.}
$$
The last integral follows from the <antiderivative> $\arctan x$.
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