hello
it seems not easy
since we have base and exponent
$\displaystyle f(x)=(1-x)^{ln(2x+1)}$
The exponent is defined if and only if $\displaystyle 2x+1>0$, i.e. $\displaystyle x>\frac{-1}{2}$
Additionally one needs $\displaystyle 1-x>0$, $\displaystyle 1-x=0$ and $\displaystyle ln(2x+1)>0$, Or, $\displaystyle 1-x \neq 0$ and $\displaystyle ln(2x+1)=r$ where r is a rational number with odd denominato.
i.e. $\displaystyle x<1$, $\displaystyle x=1$, or $\displaystyle x=\frac{e^r-1}{2}$ with $\displaystyle r$ as stated.
Answer: $\displaystyle (\frac{-1}{2},1]$ U {$\displaystyle \frac{e^r-1}{2}$ : r is a rational number number with odd denominator}.