Determination of the integrated rate equation for a reaction
After consulting the Lindemann mechanism, here is my attempt for an answer:

$$\frac{d[\ce{C}]}{dt}= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 = 0$$
$$[\ce{C}]^2=\frac{k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +k_2[\ce{C}]^2$$

Substituting $[\ce{C}]^2$,

$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +\frac{k_2k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_2k_1[\ce{A}]^2 + -k_3k_1[\ce{A}]^2 +k_2k_1[\ce{A}]^2}{k_2 +k_3}$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_3k_1[\ce{A}]^2}{k_2 +k_3}$$

Let $k'=\frac{-k_3k_1}{k_2+k_3}$, then

$$\frac{d[\ce{A}]}{dt}= k'[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{[\ce{A}]^2}= k'dt$$
$$\int_{[\ce{A}]_o}^{[\ce{A}]} \frac{d[\ce{A}]}{[\ce{A}]^2}= \int_0^t k'dt$$
$$-\left(\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o}\right) = k't$$
$$\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o} = -k't$$
$$\frac{1}{[\ce{A}]}= -k't +\frac{1}{[\ce{A}]_o} $$

I wonder how is it correct when I also used the steady state assumption in my solution. How is it different from the answer posted by my professor? Which is more correct and corresponds to the principles of chemical kinetics? Why would the other one be wrong and the other correct? (That is if my current derivation is correct :) ).