 Solution

The transformation is strictly increasing on $0\le x<1$, has range $[0,\infty )$, and its inverse is $x=y/(y+3)$. Therefore the cumulative distribution function isDifferentiation gives the probability density functionIts value at zero may be assigned arbitrarily without changing the distribution. There is no atom at zero or infinity. The expression is undefined at $X=1$, but that event has probability zero under the continuous uniform distribution, so any convention there leaves the law unchanged. Finally ${\int }_0^{\infty }3/(y+3{)}^{2}dy=1$, as required.