Overview
This is an interesting circuit. Partly, because it breaks down into parts we all recognize (an RC filter, a follower, and the 1st stage which is also a fairly familiar differential amplifier.) The fly in the ointment is that the differential amplifier is getting one of its inputs from the output, which itself is being driven by the differential amplifier. In short, there's feedback. So I really enjoyed looking over the analysis already here, as well as yours.
You wrote \$V_1 = \frac{R_2}{R_1+R_2} U_o\$ and I immediately mentally noted that since \$V_1=V_2\$ that it must also be equal to \$V_2=\frac{V_s\,\cdot\,R_1+V_3\,\cdot\,R_2}{R_1+R_2}\$. Equating them at least suggests that \$R_2\cdot V_o=V_s\,\cdot\,R_1+V_3\,\cdot\,R_2\$.
Using Solver
In any case, I found that to be a distraction and decided to just sit down and let the algebra flow out, leaving the gnarly algebra to SymPy:
eq1 = Eq( v1/r1 + v1/r2, vo/r1 ) # KCL node 1 eq2 = Eq( v2/r2 + v2/r1, vi/r2 + v3/r1 ) # KCL node 2 eq3 = Eq( v3/r1 + v3/r, io1 + v2/r1 + v4/r ) # KCL node 3 eq4 = Eq( v4/r + v4/(1/s/c), v3/r ) # KCL node 4 eq5 = Eq( vo/r1, io2 + v1/r1 ) # KCL output node eq6 = Eq( v1, v2 ) # 1st opamp requirement eq7 = Eq( v4, vo ) # 2nd opamp requirement simplify( solve( [ eq1, eq2, eq3, eq4, eq5, eq6, eq7 ], [ v1, v2, v3, v4, vo, io1, io2 ] )[vo] / vi ) -r1/(c*r*r2*s) So, this tells me that the transfer function is \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\cdot\frac1s\$. So it seems, anyway.
In the time domain, I believe this would translate into \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\int_0^tV_t\:\text{d}t\$. And if we limit ourselves to a sinusoidal \$V_t=V_{_0}\cdot\cos\left(\omega \cdot t\right)\$ then it works out to \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\cdot V_{_0}\cdot\frac1{\omega}\cdot\sin\left(\omega \cdot t\right)\$.
In short, I should expect to see a voltage gain that is inversely proportional to frequency. I should also tend to see the opamps hitting their rails with very low frequencies and diminishing towards 0 at very high frequencies.
Apply Theory
Okay. That is what I did before attempting to see what LTspice would say to me. So I set down to specify some values to use and then proceeded to make some predictions from them.
(This is what I do in my engineering logbook -- always develop the theory first, make some explicit predictions from that theory, and then go test it and write down those results, as well. That way I can tell where I fail and where I succeed and this helps me learn. It also gives me a way to look back on where I have made mistakes.)
I'll use \$R_1=10\:\text{k}\$, \$R_2=30\:\text{k}\$, \$R=10\:\text{k}\$, \$C=15.9\:\text{nF}\$ and \$V_{_0}=1\:\text{V}\$.
I am predicting a gain of \$\frac{2096.43606}{\omega}\$ for this configuration.
Test Theory
I'll now wire up a schematic in LTspice. I tend to use their LT1800 because it's just a generally good (and too expensive for me) device with rail-to-rail support and a behavioral model that seems to run well for me. So I'll use that.
Let's try a frequency of \$300\:\text{Hz}\$ first. In this case, I'd predict a voltage gain of \$\approx 1.1\$:
This shows that the peak-to-peak output is \$\approx 2.2\:\text{V}\$. With an input peak-to-peak of \$2\:\text{V}\$ and a voltage gain of \$\approx 1.1\$ that seems amazingly close.
Let's try a frequency of \$600\:\text{Hz}\$ next. In this case, I'd predict half the earlier voltage gain, or \$\approx 0.55\$:
And here we see the peak-to-peak output is \$\approx 1.1\:\text{V}\$. As expected.
One final try at a frequency of \$1\:\text{kHz}\$. In this case, I'd predict a voltage gain of \$\approx \frac13\$:
And there it is; the peak-to-peak output is \$\approx 650\:\text{mV}\$. As expected.
My logbook is complete on this, I guess.