Assume that $$\begin{equation*} p(y)=\exp{(-y)}, \end{equation*}$$ which is the exponential distribution, important for the analysis of e.g., radioactive decay. Again, $p(x)$ is given by the uniform distribution with $x \in [0,1]$, and with the assumption that the probability is conserved we have $$\begin{equation*} p(y)dy=\exp{(-y)}dy=dx, \end{equation*}$$ which yields after integration $$\begin{equation*} x(y)=P(y)=\int_0^y \exp{(-y')}dy'=1-\exp{(-y)}, \end{equation*}$$ or $$\begin{equation*} y(x)=-\ln{(1-x)}. \end{equation*}$$