If $X \sim \mathsf{Unif}(0,1)$, find the CDF of $Y = \ln\left(\frac{X}{1-X}\right)$ Assume the random variable $X$ follows distribution $\mathsf{Unif}(0,1)$ and define the new random variable $Y = \ln\left(\frac{X}{1-X}\right)$. 
Find CDF of $Y.$

My attempt. 
$F(X \leq x) = \int_{0}^{x} \ln(\frac{t}{1-t})\,dt = \ln(1-x) + x\left(\frac{x}{1-x}\right).$
Thus, 
$$
F(x) =\begin{cases} 
      0 & x < 0 \\
      \ln(1-x) + x\left(\frac{x}{1-x}\right) & 0\leq x 
\leq 1 \\
      1 & x > 1
\end{cases}
$$
Would this be right?
 A: $\mathbb{P}(Y\leq y)=\mathbb{P}(\log(\frac{X}{1-X})\leq y)=\mathbb{P}(\frac{X}{1-X}\leq e^y)=\mathbb{P}(X\leq\frac{e^y}{1+e^y})=\frac{e^y}{1+e^y}$ 
since $0<\frac{e^y}{1+e^y}<1$
A: The cdf of $Y$ is $F_Y(y) = P(Y \le y)$. I know you want to integrate something, but you can't yet because you don't even know the pdf of $Y$. You have to rewrite in the form:
$$P(Y \le y) = P(X \le h(y))$$
Then you can proceed as follows
$$P(X \le h(y)) = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx$$
$$ = \int_{\mathbb R} 1_{(-\infty,h(y))}(x) 1_{(0,1)}(x) dx = \int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx$$
And I think you know how to do the intersection based on what you tried, but I'll spell it out for you anyhoo:
$$(-\infty,h(y)) \cap (0,1) = \emptyset, h(y)\le 0$$
$$(-\infty,h(y)) \cap (0,1) = (0,h(y)), h(y) \in (0,1)$$
$$(-\infty,h(y)) \cap (0,1) = (0,1), h(y) \ge 1$$
Now for the $h(y)$:
$$F_Y(y) = {P}(Y\le y)={P}(\ln(\frac{X}{1-X})\le y)={P}(\frac{X}{1-X}\le e^y)={P}(X\le\frac{e^y}{1+e^y})$$
$$=\int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx, h(y)=\frac{e^y}{1+e^y} \in (0,1)$$
$$\therefore, F_Y(y) = {P}(Y\le y) = \int_{\mathbb R} 1_{(0,\frac{e^y}{1+e^y})}(x) dx=\frac{e^y}{1+e^y}$$
Observe that all our results hold for $y \in \mathbb R$.
