Opamp: determine the transfer function with negative RC feedback
General case

Assuming the voltage at the internal node is \$V\$

\$
\frac{0-V}{R_4}
+\frac{V_{\text{in}}-V}{R_2}
+\frac{V_{\text{out}}-V}{R_3}
+\frac{V_{\text{out}}-V}{1/(Cs)}=0\$

which can be solved for \$V_\text{out}\$

\$V_{\text{out}}(\frac{1}{R_3}+C s)=-\frac{V_\text{in}}{R_2}+(\frac{1}{R_4}+\frac{1}{R_2}+\frac{1}{R_3}+Cs)V \$

\$V_\text{out} = -\frac{R_3}{R_2 \left(C R_3 s+1\right)} V_\text{in}+
\frac{R_2 R_3+R_4 R_3+R_2 R_4+C R_2 R_4 R_3 s}{R_2 R_4 \left(C R_3
   s+1\right)} V\$

This is essentially a system with two inputs. The transfer function for \$\frac{V_\text{out}}{V_\text{in}}\$ is 

\$-\frac{R_3}{R_2 \left(C R_3 s+1\right)}\$

Case when \$R_2\$ is neglected

The node equation now is 

\$\frac{0-V_\text{in}}{R_4}
+\frac{V_{\text{out}}-V_\text{in}}{R_3}
+\frac{V_{\text{out}}-V_\text{in}}{1/(Cs)}=0\$

In this case solving for \$\frac{V_\text{out}}{V_\text{in}}\$ we get 

\$\frac{C R_4 R_3 s+R_3+R_4}{R_4 \left(\mathcal{C} R_3 s+1\right)}\$