Let $X$ be a random variable with Cauchy distribution with its probability density function $$f(x)=\frac{1}{\pi(1+x^2)}.$$ How can I find the probability density function of $$Y=\frac{2X}{1-X^2}?$$
All the problems I came across so far didn't involve rational function where both numerator and denominator were functions of $X$.
We can find the cumulative distribution function of $Y$ as:
$$F_Y(Y)=P(Y<y)=P\bigg(\frac{2X}{1-X^2}<y\bigg)$$
But I'm not sure how I can continue from there and express $F_Y(y)$ in terms of $F_X(x)$.
Hint: Use $$\frac{2X}{1-X^2} = y \iff yX^2+2X-y=0\iff X =\frac{-1\pm\sqrt{1+y^2}}{y}$$ to find the regions where $$\frac{2X}{1-X^2} \le y.$$