Please let me know where my mistake could be.
I've verified the integral $$\int_{-\infty}^\infty \frac{dt}{(t^2+1)(t^2+4)}$$ to be equal to $\frac{\pi}{6}$ with a computer math system.
However, I'm not getting the same answer when using theory of integration over contours in the complex plane.
Let $f(z):=\frac{1}{(z^2+1)(z^2+4)}$, $f(t):=\frac{1}{(t^2+1)(t^2+4)}$. The singularities of $f(z)$ are $\pm i, \pm 2i$. Let $R>2$ be the radius of the semicircle $C_R$ in the upper complex half-plane, so that $C_R$ encloses the two singularities $i$ and $2i$.
Thus we can write:
$$ \int_{-R}^R f(t)dt + \int_{C_R}f(z)dz=2\pi i (B_i + B_{2i})$$
where $B_k$ are the residues at the corresponding points.
One can determine that $(B_i + B_{2i}) = \frac{i}{4}$ and
$$ \int_{-R}^R f(t)dt = 2\pi i \frac{i}{4} + 0=-\frac{\pi}{2}$$
Please let me know where I'm mistaken.
Since $$ f(z)=\frac{1}{(z-i)(z+i)(z-2i)(z+2i)}$$ it follows that $$ \mathrm{Res}_{z=i}f(z)=\lim_{z\to i}(z-i)f(z)=\frac{1}{(2i)(-i)(3i)}=\frac{1}{6i}=-\frac{i}{6}$$ and similarly $$ \mathrm{Res}_{z=2i}f(z)=\frac{1}{i(3i)(4i)}=\frac{1}{-12i}=\frac{i}{12} $$ Therefore $$ 2\pi i\Big[\mathrm{Res}_{z=i}f(z)+\mathrm{Res}_{z=2i}f(z)\Big]=2\pi i\Big(-\frac{i}{6}+\frac{i}{12}\Big)=\frac{\pi}{6}$$