physical chemistry - Determination of the integrated rate equation for a reaction

We were given this equation to determine its integrated rate equation:

$$\ce {A + B <=>[$k_1$][$k_2$] C +D ->[$k_3$] E + F}$$

We were given these assumptions to simplify the derivation:

1. 
   

   $\displaystyle \frac{\mathrm d[\ce{A}]}{\mathrm dt} = \frac{\mathrm d[\ce{B}]}{\mathrm dt}$, and $\displaystyle \frac{\mathrm d[\ce{C}]}{\mathrm dt} = \frac{\mathrm d[\ce{D}]}{\mathrm dt}$

   
   


2. 
   

   At any time, $[\ce{A}]=[\ce{B}]$, and $[\ce{C}]=[\ce{D}]$, and $[\ce{E}] = [\ce{A}]+[\ce{B}]$

   
   



3. 
   

   $\ce{F}$ is continuously removed from the system such that $[\ce{F}] = 0$ at any given time

   
   

The answer given to us was:

$$\ln[\ce{A}] = \frac{1}{2} \ln \left( \frac{k_2+k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2 [\ce{A}]\right)$$

However, I refuse to believe that an integrated rate equation can be without the time variable. Our teacher argued that the time variable was removed by using steady state approximation, and that $\frac{\mathrm d[\ce{C}]}{\mathrm dt} =0$. However, I think there is something wrong with using the rate of formation of an intermediate in deriving the integrated rate equation for a certain reaction. Can anyone please help me?

Edit: Here is how our professor got the answer:

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt}=\frac{\mathrm d[\ce{D}]}{\mathrm dt} &= k_1[\ce{A}][\ce{B}] - k_2[\ce{C}][\ce{D}] - k_3[\ce{C}][\ce{D}] \\ \frac{\mathrm d[\ce{C}]}{dt} &= k_1[\ce{A}][\ce{B}] - (k_2 +k_3 )[\ce{C}][\ce{D}] \end{align}$$

$\ce{A}$ and $\ce{B}$ are the only species initially present such that:

$$[\ce{A}]_0 + [\ce{B}]_0 = [\ce{A}] + [\ce{B}] + [\ce{C}] + [\ce{D}] + [\ce{E}] + [\ce{F}] $$

Since $[\ce{C}] = [\ce{D}]$ and $[\ce{A}] = [\ce{B}]$ and $[\ce{F}] = 0$ and $[\ce{E}] = [\ce{A}] + [\ce{B}]$,

$$\begin{align} [\ce{A}]_0 + [\ce{B}]_0 &= 4[\ce{A}] + 2[\ce{C}] \\ [\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}] &= 2[\ce{C}] \\ \frac{[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}]}{2} &= [\ce{C}] \end{align}$$

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 \\ \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_o}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 \end{align}$$

By steady state approximation (SSA), $\frac{\mathrm d[\ce{C}]}{\mathrm dt} = 0$

$$\begin{align} \frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 = 0 \\ k_1[\ce{A}]^2 &= (k_2 +k_3)\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 \end{align}$$

Applying natural logarithms to both sides,

$$\begin{align} 2 \ln(k_1) + 2 \ln[\ce{A}] &= \ln(k_2 +k_3)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\ 2 \ln[\ce{A}] &= \ln(k_2 +k_3 ) - \ln(k_1)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\ \ln[\ce{A}] &= \frac{1}{2}\ln\left(\frac{k_2 +k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \end{align}$$

As you can see from the solution, he was able to eliminate time from the final equation by using the steady state approximation. Is this a valid move to get the integrated rate equation? Please help.