It is a great challenge to find a new explanation for such a legendary circuit because everyone knows what an op-amp integrator is. But to know a specific circuit solution does not mean that you really understand it. To (deeply) understand a circuit means something more - to see the general idea behind it that links many specific circuit implementations (op-amp, BJT, FET, tube…) You can see it even in life in the form of many non-electrical applications...
1. Op-amp inverting integrator. The idea behind this circuit solution is extremely simple and intuitive. It may sound paradoxical... but to see it you only need to remove the symbol of the ground from the circuit diagram. As you can see in Fig. 1, I have only labeled the place of the virtual ground (1) and the place of the real ground (2)... and I have no longer used these names. You understand that there is no virtual ground because there is no real ground. But if you still miss the virtual ground, then you can talk about a virtual short between node 1 and 2.
Fig. 1. Op-amp inverting integrator (only the negative power supply V- is explicitly shown)
The current path is crucial here to see the great idea. Since the input voltage is positive, the op-amp output voltage is negative and the current enters the op-amp output... then passes through the negative power supply V- and returns to the input source. The positive source V+ is not essential in this case; so it is only hinted.
2. Electric equivalent circuit. The main question to be answered is, "What does the op-amp do here?" You know that it keeps almost zero voltage between its inputs so its output voltage is always equal to the voltage drop across the capacitor. So the op-amp output serves as a following voltage source. Then let's replace the op-amp with a variable voltage source VOA to simplify this electronic circuit - Fig. 2. By the way, I conducted such a real experiment in 2001 with my students in the laboratory when we used a capacitor with high capacity and zero indicator connected between 1 and 2.
Fig. 2. Electric equivalent circuit
This simple trick is enough to show the great idea behind the circuit. The voltage source VOA is connected in series to the capacitor C so that its voltage compensates the voltage drop VC across the capacitor and the voltage between the two nodes 1 and 2 is (almost) zero. So the conclusion is:
The op-amp in the circuit of the inverting amplifier compensates the voltage drop VC across the capacitor by adding equivalent voltage VOA = VC in series.
So, the key point of this explanation is adding, not amplifying. To think of the amplifier in a negative feedback circuit not as of an amplifier but rather as of something like integrator is a powerful technique for intuitive understanding and explaining such op-amp circuits. Indeed, here it seems a little strange (integrator inside integrator)... but works...
How simple is this "magic recipe"... You want to make the imperfect RC integrator perfect? Then connect a small variable "battery" with voltage VC in series to the capacitor and (the next brilliant idea) take its inverted "copy" voltage as an output. The load will consume current from this "helping" source... not from the input source (i.e., this is a buffered output).
The power of this intuitive explanation is that we can explain this sophisticated op-amp circuit to a "six year old" (Einstein)... and that will mean we understand it ourselves...
3. Virtual short. The total voltage across the network of two elements in series - a capacitor C and compensating voltage source (VOUT), is always zero. So this network behaves as a "piece of wire" that shorts the points 1 and 2 - Fig. 3. This is what the input source "sees" when "looking" through the resistor R at the op-amp input.
Fig. 3. Equivalent circuit of the output part on the right
Figuratively speaking, the op-amp output acts as a "negative capacitor". While the "positive capacitor" C subtracts its voltage VC from the input voltage source, the op-amp "negative capacitor" adds its voltage VOUT to the input voltage.