Transfer Function - Opamp
The solution you have arrived at is correct. The circuit is a practical integrator. The resistor in parallel with capacitor limits low frequency gain and minimizes variations in output. Here is a simpler and quicker solution:

Since the opamp is in inverting configuration, the transfer function is:

$$
Av = \frac{-Z2(s)}{Z1(s)}
$$
  Note that all impedances are in s-domain. Z2(s) happens to be the parallel combination of R2 and 1/sC
$$
Z2(s) =\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}
$$
$$
Z1(s) =R1
$$
$$
\frac{vo(s)}{vin(s)} = -\frac{\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}}{R1}
$$

Which on simplification reduces to:
 $$
\frac{vo(s)}{vin(s)} = -\frac{R2}{R1\cdot(1+R2Cs)} = -\frac{\frac{R2}{R1}}{1+R2Cs}
$$

So the gain at low frequencies is -R2/R1 which without R2 would have quickly rolled off.