The first thing you need to do when trying to understand an unfamiliar circuit is not to analyze it blindly but to try to see some familiar (sub)circuits in it.

First, the superposition principle can help as to see an inverting amplifier (assuming Vp = 0) and non-inverting amplifier (Vi = 0).

Then, combining them, we can see an unfinished op-amp differential amplifier... in the sense that it has slightly unequal input gains for the inverting input (Rf/Ri) and non-inverting input (Rf/Ri + 1). It is interesting why the gain of the non-inverting input is one unit greater than that of the inverting input... but this is another topic.

There are two options to align the two input gains - by reducing the gain with 1 of the non-inverting input or by increasing the gain with 1 of the inverting input. The first requires an attenuator and the second - an amplifier. Of course, we choose the simpler solution and connect an "inverted" voltage divider with gain of Rf/(Ri + Rf) before the non-inverting input of the op-amp. Thus the non-inverting gain becomes Rf/(Ri + Rf) x (Ri + Rf)/Ri = Rf/Ri and becomes equal to the inverting gain.

The first thing you need to do when trying to understand an unfamiliar circuit is not to analyze it blindly but to try to see some familiar (sub)circuits in it.

First, the superposition principle can help as to see an inverting amplifier (assuming Vp = 0) and non-inverting amplifier (Vi = 0).

Then, combining them, we can see an unfinished op-amp differential amplifier... in the sense that it has slightly unequal input gains for the inverting input (Rf/Ri) and non-inverting input (Rf/Ri + 1). It is interesting why the gain of the non-inverting input is one unit greater than that of the inverting input... but this is another topic.

There are two options to align the two input gains - by reducing the gain with 1 of the non-inverting input or by increasing the gain with 1 of the inverting input. The first requires an attenuator and the second - an amplifier. Of course, we choose the simpler solution and connect a correcting voltage divider with gain of Rf/(Ri + Rf) before the non-inverting input of the op-amp. Thus the non-inverting gain becomes Rf/(Ri + Rf) x (Ri + Rf)/Ri = Rf/Ri and becomes equal to the inverting gain.