Find the Fourier transform of $u(x) = \frac{x \cos(2x)}{(1+x^2)^2}$ 
Find the Fourier transform of $$u(x) = \frac{x \cos(2x)}{(1+x^2)^2}$$

My work
Okay so we want $$\int_\mathbb R \frac{e^{-ixt}x\cos(2x)}{(1+x^2)^2}dx$$
Of course we want to apply the residue theorem on the function $\displaystyle f(z) = \frac{e^{-izt}z\cos(2z)}{(1+z^2)^2}$
But on which path?
I thought; let $z = \rho (\cos \theta + i \sin \theta)$, we know  that we will want eventually $\rho \to \infty$ to find our integral over $\mathbb R$.
But $$|f(z)| \le \frac{e^{(t-2)\rho\sin\theta} + e^{(t+2)\rho\sin\theta}}{2\rho^3}$$ 
So if $t < -2$, we can integrate over the positive semicircle as we see that $|f(z)| \to 0$
If $t > 2$, we can integrate over the negative semicircle; again $|f(z)| \to 0$.
But if $-2 < t < 2$? Circles are out of the question; our only hope is to maintain $\rho \sin \theta = \text{Im }(z)$ bounded, so the function still tends to $0$ as those exponential are bounded.
So I thought: let's take the rectangle $[-R, R] \times [0, \frac 12]$. The contour integral will be $0$.
I think I also know how to show that it vanishes over the vertical lines, but I'm having trouble in calculating the upper integral.
My questions
1) Is this line of thinking useful? I mean looking when the modulus of $f$ vanishes to find proper path over which integrating
2) How to finish the exercise? Integrating on the upper part seems a mess.
3) Are there easier ways to find this Fourier transform?
 A: You need to consider the symmetry of the integrand.  The FT is
$$\int_{-\infty}^{\infty} dx \frac{x \cos{2 x}}{(1+x^2)^2}  \, e^{i k x} $$
The real part of the integrand is odd, so the FT of the real part is zero.  Thus the FT is imaginary:
$$i \int_{-\infty}^{\infty} dx \frac{x }{(1+x^2)^2}  \,\cos{2 x} \sin{k x} $$
which we may rewrite as
$$\frac{i}{2} \int_{-\infty}^{\infty} dx \frac{x }{(1+x^2)^2}  \left [\sin{(k+2) x}+\sin{(k-2) x} \right ]$$
So now consider the contour integral
$$\oint_C dz \frac{z}{(1+z^2)^2} e^{i \kappa z} $$
where $\kappa = k\pm 2$, and $C$ is a semicircular contour of radius $R$ either in the upper or lower half plane according to the sign of $\kappa$.  For example, for $\kappa \gt 0$, we close the contour in the upper half plane, so the contour integral is
$$\int_{-R}^R dx \frac{x}{(1+x^2)^2} e^{i \kappa x} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{R \, e^{i \theta}}{(1+R^2 e^{i 2 \theta})^2} e^{-\kappa R \sin{\theta}} e^{i \kappa R \cos{\theta}} $$
As $R \to \infty$, the second integral is bounded by
$$\frac{2 R^2}{(R^2-1)^2} \int_0^{\pi/2} d\theta \, e^{-(2/\pi) \kappa R \theta} = \frac{\pi R}{\kappa (R^2-1)^2} \left (1-e^{-\kappa R} \right )$$
which indeed vanishes as $\pi/R^3$ in this limit.
On the other hand, the integral is equal to $i 2 \pi$ times the residue at the pole $z=i$.  Thus we have
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{x}{(1+x^2)^2} e^{i \kappa x} &= i 2 \pi \left [\frac{d}{dz} \frac{z\, e^{i \kappa z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{(1+i \kappa z) e^{i \kappa z}}{(z+i)^2} - \frac{2 z \, e^{i \kappa z}}{(z+i)^3} \right ]_{z=i} \\ &= i 2 \pi \frac14 \kappa \, e^{-\kappa}\end{align}$$
When $\kappa \lt 0$, we close in the lower half plane, and the analysis is the same, save for the fact that, because one traverses the contour clockwise, the contour integral is $-i 2 \pi$ time the residue at the pole $z=-i$.  The result is that
$$\int_{-\infty}^{\infty} dx \frac{x}{(1+x^2)^2} \sin{\kappa x} = \frac{\pi}{2} \kappa \, e^{-|\kappa|}$$
(This is the imaginary part of the FT above.)  Thus, we finally have

$$\int_{-\infty}^{\infty} dx \frac{x \cos{2 x}}{(1+x^2)^2}  \, e^{i k x} = i \frac{\pi}{4} \left [(k+2) e^{-|k+2|} + (k-2) e^{-|k-2|} \right ]$$

So to answer your questions:
1) Not really, because it is too difficult to work with the cosine in the complex integrand.  (It blows up on either semicircle as $R \to \infty$.)
2) See above.
3) In my humble opinion, not really.  Mathematica has some different way of going about it algorithmically, and it returns, as the FT,
$$\frac{1}{2} i \pi  e^{k+2} \theta (-k-2)+\frac{1}{4} i \pi  e^{k+2} k \theta (-k-2)-\frac{1}{2} i \pi  e^{k-2} \theta (2-k)+\frac{1}{4} i \pi  e^{k-2} k \theta (2-k)-\frac{1}{2} i \pi  e^{2-k} \theta (k-2)\\+\frac{1}{4} i \pi  e^{2-k} k \theta (k-2)+\frac{1}{2} i \pi  e^{-k-2} \theta (k+2)+\frac{1}{4} i \pi  e^{-k-2} k \theta (k+2)$$
I  verified that this expression is equal to the simple one I returned as the answer.  That said, I don't see how any scheme that returns this answer is any easier than what I have outlined, and even so, you are now left with simplifying this monster, which is no fun.
ADDENDUM
I just realized the OP defined the FT using $e^{-i k x}$, so the result will have $-i$ rather than $+i$ in front.
A: Since $e^{-itx}=\cos(tx)-i\sin(tx)$, then the integral can be written as
$$I=\int_{-\infty}^{\infty}\frac{x\cos(2x)\cos(tx)-ix\cos(2x)\sin(tx)}{(1+x^2)^2}\,dx$$
The real part of the integrand is an even function, hence its integral value simply equals zero by simmetry. Using trigonometric identity, then
$$I=-\frac{i}{2}\int_{-\infty}^{\infty}\frac{x\sin(t+2)x+x\sin(t-2)x}{(1+x^2)^2}\,dx$$
Let us consider a well-known integral which can be derived by using Fourier transform or complex analysis.
$$\int_{-\infty}^{\infty}\frac{\cos(bx)}{x^2+a^2}\,dx=\frac{\pi e^{-a|b|}}{a}\quad,\quad\text{for}\,a>0$$
Differentiating both sides with respect to $a$ and $b$ once, with the aid of Wolfram Alpha, we have
$$\int_{-\infty}^{\infty}\frac{2ax\sin(bx)}{(x^2+a^2)^2}\,dx=\pi b\,e^{-a|b|}\quad\Longrightarrow\quad\int_{-\infty}^{\infty}\frac{x\sin(bx)}{(x^2+a^2)^2}\,dx=\frac{\pi b\,e^{-a|b|}}{2a}$$
Hence
$$I=-\frac{i\pi}{4}\left[(t+2)\,e^{-|t+2|}+(t-2)\,e^{-|t-2|}\right]$$
