A problem on transformation of a random variable. 
$ f(x) =
\begin{cases}
1  & \text{if $0\le x \le 1$ } \\
0  & \text{otherwise }\end{cases}$

Find the PDF of Random variable $Y=\dfrac{X}{1+X}$
$F(Y\le y)=F\bigg(\dfrac{X}{1+X}\le y\bigg)=F\bigg(X\le \dfrac{y}{1-y}\bigg)$
$U=F_x\bigg(\dfrac{y}{1-y}\bigg)$
$\dfrac{dU}{dy}=f(y)\bigg(\dfrac{1}{(1-y)^2}\bigg)$
$f(y)=\bigg(\dfrac{1}{(1-y)^2}\bigg)$
Please check if I did any mistake. I am not sure how to figure out domain after transformation. Heres my try.
$0\le x \le 1$
$1\le x+1\le2$
$1\ge \dfrac{1}{x+1}\ge\dfrac{1}{2}$
$x\ge \dfrac{x}{x+1}\ge\dfrac{x}{2}$
$x\ge y\ge\dfrac{x}{2}$
Is this a correct way to approach result? 
 A: There are simple and practical reasons why one writes $F_X(x)$ and not $F_x(X)$ or $F_x(x)$ or $F_X(X).$ If one does not know about this, how would one even understand something like $\Pr(X\le x)\text{ ?}$
\begin{align}
\text{For } 0\le y \le \frac 1 2,\text{ we have } & \frac X {1+X} = 1 - \frac 1 {1+X} \le y \\[10pt]
\text{iff } & \frac 1 {1+X} \ge 1 - y \\[10pt]
\text{iff } & 1 + X \le \frac 1 {1-y} \\[10pt]
\text{iff } & X \le \frac 1 {1-y} - 1 = \frac y {1-y}. \\[10pt]
\text{Therefore for } 0 \le y \le \frac 1 2, \quad & \Pr\left( \frac X{1+X} \le y \right) = \Pr\left( X\le \frac y {1-y} \right) \\[10pt]
= {} & \int_0^{y/(1-y)} 1\, dx = \frac y {1-y}. \\[12pt]
\text{Hence } & f_Y(y) = \frac d {dy}\, \frac y {1-y} = \cdots.
\end{align}
A: 
So you are asking for the pdf 
$$F(Y\le y)=F(X\le \frac{y}{1-y})$$
so you can write 
$$F(Y\le y)=\int_0^{\frac{y}{1-y}} 1 \, dx=\frac{y}{1-y}$$
$$f_Y(y)=\frac{1}{(1-y)^2}$$
$f_Y(y)$ is your PDF
$\frac{1}{0}$ is not defined  thus   $$(1-y)^2\ne 0$$  $$1-y \ne 0$$ $$y \ne 1$$  
also, you want $$0\le\frac{y}{1-y}\le 1$$ Solve it to get your domain
