How to do the Fourier Transform of bounded function?
Well, imho, one good way to do it is to take the signal's derivative in time and make use of the property $$\frac{d}{dt}x(t) \longleftrightarrow j2\pi f X(f)$$

I'll try to show it analytically.

Since the signal is discontinuous at $t=\pm 2$ you should include Delta function in the derivative, that is, $$\frac{d}{dt}\Big[\frac{A}{2}t\:\mathrm{rect}\Big(\frac{t}{4}\Big)\Big]= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big)$$

Given that a unitary rectangular pulse can be written as a sum of two unit step functions $$\mathrm{rect}\Big(\frac{t}{T}\Big) = u\Big(t+\frac{T}{2}\Big) - u\Big(t-\frac{T}{2}\Big)$$ then the pulse's derivative is $$u'\Big(t+\frac{T}{2}\Big) - u'\Big(t-\frac{T}{2}\Big) = \delta\Big(t+\frac{T}{2}\Big) - \delta\Big(t-\frac{T}{2}\Big)$$

Going back, we get $$\frac{d}{dt}x(t)= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big) = \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\Big(\delta(t+2) - \delta(t-2)\Big)$$

The sampling property of a Delta function yields

$$t\delta(t \pm t_0) = t\Big|_{t=\pm t_0}\delta(t \pm t_0) = (\mp t_0)\delta(t\pm t_0)$$

and in your case

$$\frac{A}{2}t\Big(\delta(t+2) - \delta(t-2)\Big)=-A\delta(t + 2)-A\delta(t - 2) $$

and in total 

$$\frac{d}{dt}x(t)= \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) + \frac{A}{2}t\:\mathrm{rect}'\Big(\frac{t}{4}\Big) = \frac{A}{2}\mathrm{rect}\Big(\frac{t}{4}\Big) -A\delta(t + 2)-A\delta(t - 2)$$

The Fourier Transform of the latter is 

$$F\Big\{\frac{d}{dt}x(t)\Big\} = 2A\mathrm{sinc}(4f) - Ae^{j2\pi 2f} - Ae^{-j2\pi 2f} = 2A\mathrm{sinc}(4f) - 2A\cos(2\pi 2f)$$

and from the derivative property $$X(f) = \frac{1}{j2\pi f}\Big(2A\mathrm{sinc}(4f) - 2A\cos(2\pi 2f)\Big) = \frac{jA}{2\pi f}\Big(2\cos(2\pi 2f) - 2\mathrm{sinc}(4f) \Big)$$

which gives 

$$X(f) = jA\Big(\frac{2\cos(2\pi 2f)}{2\pi f} - \frac{\sin(4\pi f)}{(2\pi f)^2} \Big)$$

Finally, with the $j\omega$ notation

$$X(j\omega) = jA\Big(\frac{2\omega\cos(2\omega)}{\omega^2} - \frac{\sin(2\omega)}{\omega^2} \Big) =jA\Big(\frac{2\omega\cos(2\omega) - \sin(2\omega)}{\omega^2} \Big)$$