= Solution
For a pole of order $N$ at $z_0$, the <residue> is $\operatorname{Res}_{z_0}F=\frac1{(N-1)!}\lim_{z\to z_0}\frac{d^{N-1}}{dz^{N-1}}[(z-z_0)^NF(z)]$. For the rational integral, close the real-axis contour with an upper semicircle. The poles are double at $i$ and simple at $2i$, with
$$
\operatorname{Res}_iF=\left.\frac d{dz}\frac1{(z+i)^2(z^2+4)}\right|_{z=i}=-\frac i{36},\qquad \operatorname{Res}_{2i}F=\frac1{(4i)(-3)^2}=-\frac i{36}.
$$
The arc integral vanishes as $O(R^{-5})$. By the <residue theorem> the whole real-axis integral is $\pi/9$, and evenness gives
$$
\boxed{\int_0^\infty\frac{dx}{(x^2+1)^2(x^2+4)}=\frac\pi{18}.}
$$
For the second calculation, reciprocal expansion of the sine gives $1/\sin z=z^{-1}+z/6+O(z^3)$. Multiplying by $\cos z=1-z^2/2+O(z^4)$ yields $\cot z=z^{-1}-z/3+O(z^3)$. Therefore
$$
\frac{\pi\cot\pi z}{z^2}=z^{-3}-\frac{\pi^2}{3}z^{-1}+O(z),\qquad\boxed{\operatorname{Res}_0=-\frac{\pi^2}{3}.}
$$
At every nonzero integer $n$, $\pi\cot\pi z$ has a simple pole with residue one, giving residue $1/n^2$ after division by $z^2$.
The <cotangent contour summation> uses the positively oriented square of half-side $R=N+1/2$. Its length is $8R$ and $|z|\geq R$ on its boundary. The supplied cotangent bound implies $|\oint Fdz|\leq8\pi C/R\to0$. Its enclosed residues sum to $-\pi^2/3+2\sum_{n=1}^Nn^{-2}$. The <residue theorem> therefore proves the <Basel problem> value
$$
\boxed{\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}.}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2-residue-contours.png]
{title=Upper-half-plane and square contours for the rational integral and the Basel sum}
{description=Original contour diagrams. Both are counterclockwise. The upper semicircle encloses poles at $i$ and $2i$; the example square has half-side $5/2$, enclosing the integer poles from $-2$ to $2$.}
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