Residue integral problem Have this Integral. 
$$\int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}\,dx$$
Been working on similar problems but the cos bother me in this problem. Can anyone help me get started? What should I do first?
I´m given that i could use 
$$\operatorname{Re}{\left ( \int_{-\infty }^\infty \frac{e^{2ix}}{(x^2+1)(x^2+4)^2} \, dx \right )} = \int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}dx$$
How can the real part of this integral help me?

Very much like the example II in this wiki page, we write:
$$
   \int_C \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2} \mathrm{d} z = 
   \int_{-\infty}^\infty \frac{\mathrm{e}^{2 i x}}{(x^2+1)(x^2+4)^2} \mathrm{d} x + \int_{0}^{\pi} \frac{\exp(2 i R \mathrm{e}^{i \varphi} )}{((R \mathrm{e}^{i \varphi})^2+1)((R \mathrm{e}^{i \varphi})^2+4)^2}  i R \mathrm{e}^{i \varphi} \mathrm{d} \varphi
$$
The latter integral vanishes as $R \to + \infty$ because 
$$ \begin{multline}
 \lim_{R \to + \infty} \operatorname{abs}\left( \frac{\exp(2 i R \mathrm{e}^{i \varphi} )}{((R \mathrm{e}^{i \varphi})^2+1)((R \mathrm{e}^{i \varphi})^2+4)^2}  i R \mathrm{e}^{i \varphi} \right) =\\
  \lim_{R \to + \infty} \frac{R \exp(-2 R \sin(\varphi))}{\sqrt{1+2 R^2 \cos(2 \varphi) + R^4} \left( 16 + 8 R^2  \cos(2 \varphi)  + R^4 \right)} = 0
\end{multline}
$$
The contour integral is evaluated by residues. Let $f(z) = \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2}$. Then
$$
 \begin{eqnarray}
\int_C \frac{\mathrm{e}^{2 i z}}{(z^2+1)(z^2+4)^2} \mathrm{d} z &=&
   2 \pi i \left( \operatorname{Res}_{z=i} f(z) + 
    \operatorname{Res}_{z=2 i} f(z) \right ) \\
  &=& 2 \pi i \left(  \left. (z-i)f(z) \right|_{z=i} + \left. \frac{\mathrm{d}}{\mathrm{d} z} (z-2 i)^2 f(z) \right|_{z=2i} \right) \\ 
  &=& \frac{\pi}{9 \mathrm{e}^2} - \frac{23 \pi}{144 \mathrm{e}^4}
 \end{eqnarray}
$$
Since the result is real, it is also the value of the original integral:
$$
  \int_{-\infty}^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2} \mathrm{d} x = 
     \frac{\pi}{9 \mathrm{e}^2} - \frac{23 \pi}{144 \mathrm{e}^4} \approx 0.038
$$
