# Pushforward Measure of $\frac{x}{x^2+1}$

I was doing some exercices that was offered during the lecture and came across very interesting one. I didn't make it, then asked my teacher and didn't understand what He tried to explain me. So here it is:

I have to explain the probability distribution function $F_\nu(t)=\nu((-\infty,t])=\mu(f^{-1}(-\infty,t])$ in terms of function $F_\mu(t)$. Where $f(x)=\frac{x}{x^2+1}$.

For example if $f(x)=x^2$ then $F_\nu(t)=0$ if $t<0$ and $F_\nu(t)=F_{\mu}(\sqrt{t}) - F_\mu(-\sqrt{t})$ when $t\geq 0$.

I think that it would be very helpful to find an inverse function. I don't know if it exists, but I remember that during the calculus when we had to change variables and we had functions like $\sqrt{2x-x^2}$ there was an inverse function on some interval and it was given like $1+\sqrt{1-y^2}$. I don't know if it's right, but I think I need something like this in this situations.

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Via calculus you find that the value of $x/(1+x^2)$ is maximized by $x=1$, at which point $x/(1+x^2)=1/2$, and minimzed by $x=-1$. So $-1/2 \le x/(1+x^2)\le 1/2$ for all real $x$.
Since $1+x^2$ is positive for all $x$, the inequality $x/(1+x^2)\le t$ is equivalent to $x\le t+tx^2$. Completing the square, you find that $x=t+tx^2$ precisely if $x=\dfrac{1\pm\sqrt{1-4t^2}}{2t}$. Here we rely on the fact that $-1/2\le t\le1/2$, since otherwise the expression under the radical is negative.
If $1/2\ge t>0$, we get $x/(1+x^2)\le t$ if and only if either $x\le\dfrac{1-\sqrt{1-4t^2}}{2t}$ or $x\ge\dfrac{1+\sqrt{1-4t^2}}{2t}$.
Then do something similar for the case $-1/2\le t<0$.
If $t=0$, then $x/(1+x^2)\le t$ iff $x\le 0$.