I think it is lots of algebra. I think you can write some nodal analaysis equations and solve. Assuming this is an ideal opamp, the (-) node of the opamp between R2 and R3 is Vin, so mark that node as Vin and mark the top of R1 as Vx. Then write node equations
sum of current out of the Vin node =0
(Note: current into (-) node of opamp is zero, ideal opamp assumption).
(Vin-Vx)/(R3 + 1/sC1) + (Vin-Vo)/R2 = 0 (Kirchoff's current law)
sum of current out of the Vx node =0
(Vx-Vin)/(R3 + 1/sC1) + Vx/R1 + (Vx-Vout)/(1/sC2) =0
Now eliminate Vx from the equations and you will have the expression of Vo in terms of Vin. How do you eliminate Vx? Take the first equation and solve it for Vx and you will get
Vx = Vin + ( (R3 + 1/sC1)/R2 )*(Vin-Vo)
substitute this in for Vx in the second equation (now you should have only Vin and Vout).
Your resulting equation should be a second order equation.
Does this help?
Best regards,
v_c
Added after 1 hours 13 minutes:
with a little algebra
it can be shown that
\[ \frac{V_{out}(s)}{V_{in}(s)}= \frac{1 + (C1 R3 + C1 R2 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}{1 + (C1 R3 + C1 R1 + C2 R1) s + (C2 C1 R1 R3 + C2 C1 R1 R2) s^2}\]
Now this is a standard form of the second order system where the denominator is of form \[1 + (\frac{s}{\omega_0}) \frac{1}{Q} + (\frac{s}{\omega_0})^2\] and \[\omega_0 = 2 \pi f_0\]. So looking at denominator of that equation we see that the term multiplying \[s^2\] must be \[\frac{1}{\omega_0^2}\] so
\[\omega_0^2 = \frac{1}{C2 C1 R1 R3 + C2 C1 R1 R2}=\frac{1}{C2 C1 R1 (R3 + R2)} \] so we get \[\omega_0 = \frac{1}{\sqrt{C2 C1 R1 (R3 + R2)}} \] and \[f_0 = \frac{\omega_0}{2 \pi} = \frac{1}{2 \pi \sqrt{C2 C1 R1 (R3 + R2)}} \]
OK now the term that is multiplying the s term in the denominator must be \[\frac{1}{\omega_0 Q}\] so equating we get
\[\frac{1}{\omega_0 Q}=C1 R3 + C1 R1 + C2 R1\] solving for \[Q\] and noting the value of \[\omega_0\] that we just solved for we get \[Q=\frac{\sqrt{C2 C1 R1 (R3 + R2)}}{C1 R3 + C1 R1 + C2 R1}\]. I think to get the value of \[Q\] you have to assume \[R3\] is small. See if your datasheet does use this approximation. So assuming R3 is much much smaller than other values (limit as R3 goes to zero), you get.
\[Q=\frac{\sqrt{C2 C1 R1 R2}}{C1 R1 + C2 R1}=\frac{1}{C1+C2} \sqrt{ \frac{C2 C1 R2}{R1} }\]
Now the final value is the midband gain. Note that both of the terms multiplying \[s^2\] on the numerator and the denominator is the same they are both \[1/\omega_0^2\], which is the resonant frequency. If the transfer function when evaluated at this frequency at \[s=j \omega_0\], the \[s^2\] terms will cancel with 1, since \[ (j \omega_0/\omega_0)^2 = -1\]. At the resonant frequency the gain is just given by the ratio of the factors that multiply the s term. That is just given by
\[Gv = \frac{C1 R3 + C1 R2 + C1 R1 + C2 R1}{C1 R3 + C1 R1 + C2 R1}\]
Now once again your datasheet must have an approximation, this time they must be assuming C1=C2, because if you do that the capacitance terms cancel from the top and the bottom of the expressions and you get.
\[Gv = \frac{R3 + R2 + R1 + R1}{R3 + R1 + R1} = \frac{R3 + R2 + 2 R1}{R3 + 2 R1} \]. Now if you divide top and bottom by R1 you get
\[Gv = \frac{\frac{R3 + R2}{R1} + 2 }{\frac{R3}{R1} + 2} \]. Now all you have to do is take 20 times the log of this and you have the gain in dB.
I think that is all I have to say ... lots of work for just 3pts
Best regards,
v_c