Solving $\frac{dx}{dt} = A \frac{ (1-x)}{(t-t^2)} - \frac{(B*x -C*x^2)}{(t-t^2 )*(t-x)}$ (using wolfram/mathematica)

Let $u=t-x$ ,

Then $x=t-u$

$\dfrac{dx}{dt}=1-\dfrac{du}{dt}$

$\therefore1-\dfrac{du}{dt}=\dfrac{A(1-t+u)}{t-t^2}-\dfrac{B(t-u)-C(t-u)^2}{(t-t^2)u}$

$u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{Au^2}{t-t^2}+\dfrac{Cu^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$

$u-u\dfrac{du}{dt}=\dfrac{Au}{t}+\dfrac{(A+C)u^2+(B-2Ct)u+Ct^2-Bt}{t-t^2}$

$u\dfrac{du}{dt}=\dfrac{(A+C)u^2}{t^2-t}+\left(\dfrac{B-2Ct}{t^2-t}-\dfrac{A}{t}+1\right)u+\dfrac{Ct^2-Bt}{t^2-t}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dt}=-\dfrac{1}{v^2}\dfrac{dv}{dt}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dt}=\dfrac{A+C}{(t^2-t)v^2}+\left(\dfrac{B-2Ct}{t^2-t}-\dfrac{A}{t}+1\right)\dfrac{1}{v}+\dfrac{Ct^2-Bt}{t^2-t}$

$\dfrac{dv}{dt}=-\dfrac{(Ct^2-Bt)v^3}{t^2-t}-\left(\dfrac{B-2Ct}{t^2-t}-\dfrac{A}{t}+1\right)v^2-\dfrac{(A+C)v}{t^2-t}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

I tried to use Mathematica to solve your problem and it solved pretty fast. Here is what I've done.