How this voltage resonant soft-switching IGBT works?
If you extracted the full picture from your link it would impart more information (I have added the definition of FWD and written it in purple so as not to confuse). I've also re-arranged it to make it easier to follow: -




Inductor current ramps upwards when IGBT turns on (as per \$\dfrac{di}{dt} = \dfrac{V}{L}\$). It's slope starts to tail off due to C1 i.e. it isn't a true linear rise like a pure inductor.
IGBT turns off and residual energy in the inductor forms a tuned circuit with C1 
You only get half a cycle of resonance because the diode across the IGBT shunts the negative portion of the sine wave and the IGBT turns on again before it can rise positively again.


In all of the above waveforms, residual voltages like the IGBT on voltage and the forward conduction voltage of the FWD are ignored. They are ignored because they represent a small fraction of the overall waveform voltages.

They shouldn't be ignored when making conduction loss calculations though.