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