contour integration - Complex Analysis - $int_0^inftyfrac{cos(5x)}{(1+x^2)^2}mathrm{d}x$

How to calculate the following integral using complex analysis?
$\int_0^\infty\frac{\cos(5x)}{(1+x^2)^2}\mathrm{d}x$.

So far I have

$$\int_0^\infty\frac{\cos(5x)}{(1+x^2)^2}\mathrm{d}x = \int_{-\infty}^\infty\frac{1}{(1+x^2)^2}e^{5ix}\mathrm{d}x$$ Then, $$Res(f(x),i)=\frac{d}{dx}[e^{5ix}]|_i=5ie^{5ix}|_i=2\pi i5ie^{5i(i)}=\frac{-10\pi}{e^5}$$
Then I might have to multiply by 1/2 to get from 0 to infinity only but that gives $\frac{-5\pi}{e^5}$ and the answer should be $\frac{3\pi}{2e^5}$ and I am not sure what I am doing wrong...

Answer

$$\int_{0}^{+\infty}\frac{\cos(5x)}{(1+x^2)^2}\,dx = \frac{1}{2}\text{Re}\int_{-\infty}^{+\infty}\frac{e^{5ix}}{(1+x^2)^2}\,dx \tag{1}$$
and $x=i$ is a double pole for $\frac{e^{5ix}}{(1+x^2)^2}$, in particular

$$\text{Res}\left(\frac{e^{5ix}}{(1+x^2)^2},x=i\right) = \lim_{x\to i}\frac{d}{dx}\left(\frac{e^{5ix}}{(x+i)^2}\right)=-\frac{3i}{2e^5}\tag{2}$$

and
$$\int_{0}^{+\infty}\frac{\cos(5x)}{(1+x^2)^2}\,dx = \text{Re}\left(\frac{(-3i)\cdot(\pi i)}{2e^5}\right)=\color{red}{\frac{3\pi}{2e^5}}.\tag{3}$$