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?

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The integral under the $\operatorname{Re}$ can be computed by method of residues – Sasha Oct 6 '11 at 22:04
No special treatment is required for the cosine. You can just substitute the pole $z_0$ into $f(z)(z-z_0)$ like you would with any other function. – joriki Oct 6 '11 at 22:09
The point of the hint is that the real part of $\int_a^b (u(t) + iv(t))\, dt$ is $\int_a^b u(t)\, dt$. – Dylan Moreland Oct 6 '11 at 22:15

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: $$\int_{-\infty}^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2} \mathrm{d} x = \frac{\pi}{9 \mathrm{e}^2} - \frac{23 \pi}{144 \mathrm{e}^4} \approx 0.038$$