How to tell if wavepacket is nonstationary vs. stationary? $$\psi(x,t) = \int_{-\infty}^{\infty} A(k)e^{i(kx-w(k)t)} \mathrm{d}k$$
Also, what properties of this equation will tell me whether or not this wave is going to spread over time?
 A: That's a general solution for a wave packet.
1) A(k) is the spatial Fourier transform of the wave packet at t = 0. Just set t=0 inside the integral and you have the inverse Fourier (or Fourier) tRansform of A(k). 
2) That means that A(k) is the spatial spectra of the initial wave packet. So the initial packet spatial extent is roughly 1/(range in k where A is nonzero, or practically non negligible). 
3) the wave packet little waves moves to the right in x with phase velocity (i.e., the velocity of the little waves inside the packet) of w(k)/k. If w(k)/k is constant but not zero, call it $v_p$, the phase velocity and the velocity of the packet as a whole, called the group velocity, $v_g$, are the same. That is called a non-dispersive packet (sometimes a packet that is traveling in a non-dispersive media), as it travels. Electromagnetic wave packets in vacuum (and no gravity) are non-dispersive. 
4) if w(k) = 0, the packet nor the wavelets inside move. It is static. 
5) if w(k)/k is not constant, the packet or media are called dispersive. The group velocity is then $v_g = dw/dk$. That's the velocity of the packet as a whole. 
6) if w(k) = constant, it is a sinusoidal (or complex exponential) oscillation in time of the initial wave packet, and it does not move to the right. You can see that from the integral you wrote. 
So it'll move to the right under cases 3 when no dispersive, and 5. Ie, non-stationary. If w(k) is either 0 or a constant the packet won't move to the right, i.e. it is stationary. For 6 it is stationary as a packet, but oscillates in place.
You can see a math description with some of those cases described and drawn for a couple different initial packets, like a Gaussian shape at
http://users.physics.harvard.edu/~schwartz/15cFiles/Lecture11-WavePackets.pdf. There's too many possibilities for both the initial packet and w(k) to show them all. The function w(k) is called the dispersion relations, and they also explain why. And you can google more on it. 
