Frequency of TL084-based oscillator producing a triangular wave
The equation you have should be correct. I derive it this way:

Note that the only assumption I'm making here is that the output swing of the first opamp is symmetrical about zero, &plusmn;\$V_{O1}\$. It doesn't matter whether this swing is related to the supply voltage or not.

The voltage \$V_{FB}\$ at the + input of the first opamp is:

$$V_{FB} = \frac{R3 \cdot V_{O1} + R2 \cdot V_{O2}}{R2 + R3}$$

Solve this for \$V_{O2}\$:

$$V_{O2} = \frac{(R2 + R3)}{R2}  V_{FB} - \frac{R3}{R2} V_{O1}$$

The circuit switches when \$V_{FB}\$ reaches 0, so this condition defines the peak-to-peak output voltage in terms of the peak-to-peak swing of the comparator:

$$V_{O2(PP)} =  - \frac{R3}{R2} V_{O1(PP)} = - \frac{R3}{R2} 2 V_{O1}$$

Now consider the integrator, whose output voltage with respect to time is:

$$V_{O2(PP)} = - \frac{V_{O1}}{R C} t$$

To get the half-period of the output waveform, set these two equations equal to each other. What this says is that the step change in \$V_{FB}\$ caused by \$V_{O1}\$ must be equal to the height of the slope change caused by \$V_{O2}\$:

$$- \frac{R3}{R2} 2 V_{O1} = - \frac{V_{O1}}{R C} t$$

Solve for t:

$$t = \frac{R3}{R2} 2 R C$$

The frequency is 1/2t:

$$f =  \frac{R2}{R3} \frac{1}{4 R C}$$