= 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> is
$$
\boxed{F_Y(y)=\begin{cases}0,&y<0,\\[2pt]\dfrac{y}{y+3},&y\ge0.\end{cases}}
$$
Differentiation gives the <probability density function>
$$
\boxed{f_Y(y)=\frac{3}{(y+3)^2}\ \ (y>0),\qquad f_Y(y)=0\ \ (y<0).}
$$
Its 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 $\int_0^\infty3/(y+3)^2\,dy=1$, as required.
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