The directions are to find the answer by using residues.
Question
$$\int_{0}^{\infty}\frac {x^2\,dx}{(x^2+1)(x^2+4)}$$
$2i, -2i, i, -i$ are the poles and the corresponding residues are $\frac {\pm1}{6i}, \frac{\pm1}{3i}$
So after proving that the integrand over the semicircle part (and not the segment $[-R, R]$ ) of the contour $C$$ $\begin{aligned} z(t) = Re^{i\theta t} &&R>2 \end{aligned} goes to 0 as $R\rightarrow\infty$ so that in effect $$\int_{\gamma}\frac {x^2\,dx}{(x^2+1)(x^2+4)} = \int_{-R}^{R}\frac {x^2\,dx}{(x^2+1)(x^2+4)}$$ where $\gamma$ is the closed semicircle. Because this function is even we would take half of this integral from $[0, \infty)$. Now, I tried to use the Cauchy residue theorem. However, as one can notice the residues all add up to 0. However, the answer by using partial fractions involves inverse tangent and the answer is $\frac{\pi}{6}$ I am confused
If your contour consists of the upper semi-circle of radius $R$ and the line segment $[-R,R]$ then there are only two poles $i$ and $2i$ inside the contour. It appears that you are including all four poles in your calculation.