[tex]\dfrac{\mathrm dx}{\mathrm dt}=(x-1)(2x-1)[/tex]
is a separable ODE, as
[tex]\dfrac{\mathrm dx}{(x-1)(2x-1)}=\mathrm dt[/tex]
Decompose the left side into partial fractions:
[tex]\dfrac1{(x-1)(2x-1)}=\dfrac1{x-1}-\dfrac2{2x-1}[/tex]
Then integrating both sides gives
[tex]\displaystyle\int\left(\frac1{x-1}-\frac2{2x-1}\right)\,\mathrm dt=\int\mathrm dt[/tex]
[tex]\ln|x-1|-\ln|2x-1|=t+C[/tex]
Solve for [tex]x(t)[/tex]:
[tex]\ln\left|\dfrac{x-1}{2x-1}\right|=t+C[/tex]
[tex]\dfrac{x-1}{2x-1}=e^{t+C}=Ce^t[/tex]
[tex]x-1=(2x-1)Ce^t[/tex]
[tex]x(1-2Ce^t)=1-Ce^t[/tex]
[tex]\implies\boxed{x(t)=\dfrac{1-Ce^t}{1-2Ce^t}}[/tex]
