By the algebra below I keep getting $F=\frac{1}{F}$, where $F$ is the log function. I appreciate if someone can point out the error.
$$F(t)=\frac{1}{1+e^{-t}}$$ $$f(t)=\frac{dF}{dt}=\frac{e^{-t}}{(1+e^{-t})^2}$$ $$\frac{f}{F}=\frac{1}{1+e^t}$$ $$\frac{f}{F}=\frac{d\ln{F}}{dt} \rightarrow \int d\ln{F}=\int \frac{1}{1+e^t} dt $$ $$\ln{F}=\ln(1+e^{-t})+C$$ $$F=1+e^{-t}$$ Looking back at the first equation, this implies $$F=\frac{1}{F}$$