$V_2= 0 ,I_B=0 \\ ∴I_C=I_B+I_f , but I_B=0 \\ ∴I_C=I_f \\ ∴I_C=C \frac{dV_c}{dt} \\ ∴I_C=C \frac{dV_c}{dt} ,V_c=V_{in}-V_2 \\ I_f=c \frac{d}{dt} (V_{in}-V_e ),I_f=\frac{V_2-V_0}{R_f} \\ \text{But} I_C=I_f \text{because} I_B=0 \\ ∴C_1 \frac{d}{dt} (V_{in}-V_2 )=\frac{V_2-V_0}{R_f} \\ \text{Using concept of virtual ground,} \\ V_1-V_2=0 \\ ∴C_1 \frac{d}{dt} V_{in}=\frac{-V_0}{R_f} \\ ∴V_0=R_f C_1 \frac{d}{dt} (V_{in} ) ∴-R_{C1} \text{times the time derivative of i/p voltage}$