PDF of $\frac{X}{1+X^2}$ in terms of the PDF of $X$ Given a r.v. $X$ with a know probability density $g$, I would like to find the probability density of $\frac{X}{1+X^2}$. Below I write my calculations:
denote 
$f(X) := \frac{X}{1+X^2}$
Now $f$ is not monotone so we can't apply the known theorem, we proceed by finding the solutions to $\frac{X}{1+X^2} < c$ for the solutions to exist we need $-1/2<c < 1/2$ and the solutions to the associated quadratic equation will be $x_{1-2} = 1 \pm\sqrt{1-4c^2}/ (2c)$.
I obtain that 
$$P(f(X) < c) = P( \{X <1 -\sqrt{1-4c^2}/ (2c)\} \cup \{ X > 1 +\sqrt{1-4c^2}/ (2c)  \}) = 1-P(X < 1 +\sqrt{1-4c^2}/ (2c)) + P(X <1 -\sqrt{1-4c^2}/ (2c))$$
So the resulting density of $\frac{X}{1+X^2}$ will be $$1+ \frac{ 4c/ \sqrt{1-4c^2} + 2(1+\sqrt{1-4c^2})}{4c^2}g(1 +\sqrt{1-4c^2}/ (2c)) + \frac{ 4c/ \sqrt{1-4c^2} - 2(1-\sqrt{1-4c^2})}{4c^2}g(1 -\sqrt{1-4c^2}/ (2c)$$
with the condition that $-1/2<c<1/2$.
Is this the correct approach?
 A: You might be interested to know that, if $X$ has PDF $f_X$ and if $Y=h(X)$ for some function $h$ regular enough then, the PDF $f_Y$ of $Y$ is given by

$$f_Y(y)=\sum_{x:h(x)=y}\frac1{|h'(x)|}f_X(x)$$

In your case, $$h(x)=\frac{x}{1+x^2}$$ hence $f_Y(y)=0$ for $|y|\geqslant\frac12$. For every $0<|y|<\frac12$, $h(x)=y$ if and only if $x=\xi_\pm(y)$, where $$\xi_\pm(y)=\frac{1\pm\sqrt{1-4y^2}}{2y}$$ Furthermore, $$h'(x)=\frac{1-x^2}{(1+x^2)^2}$$ hence $$h'(\xi_\pm(y))=\mp\frac{y\sqrt{1-4y^2}}{\xi_\pm(y)}$$ which yields, for every $0<|y|<\frac12$,
$$f_Y(y)=\sum_{\pm}\frac{\xi_\pm(y)}{y\sqrt{1-4y^2}}f_X(\xi_\pm(y))$$
The actual computational details in the specific case that interests you may be slightly involved but (I hope it is apparent that) the method itself is straightforward, even when, as here, $h$ is not injective.
A: Using the CDF
Let
$$
G(x)=\int_{-\infty}^xg(t)\,\mathrm{d}t\tag{1}
$$
be the CDF of $X$ and $Y=\frac{X}{X^2+1}$. Then the CDF of $Y$ is
$$
\begin{align}
F(\lambda)
&=P(y\lt\lambda)\\
&=P\left(\frac{x}{x^2+1}\lt\lambda\right)\\
&=\left\{\begin{array}{}
P\left(x\lt\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-P\left(x\lt\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)+1&\text{if $0\le\lambda\lt\frac12$}\\
P\left(x\lt\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-P\left(x\lt\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)&\text{if $-\frac12\le\lambda\lt0$}
\end{array}\right.\\
&=\left\{\begin{array}{}
G\left(\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-G\left(\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)+1&\text{if $0\lt\lambda\le\frac12$}\\
G(0)&\text{if $\lambda=0$}\\
G\left(\frac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)-G\left(\frac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)&\text{if $-\frac12\le\lambda\lt0$}
\end{array}\right.\tag{2}
\end{align}
$$
Then
$$
\begin{align}
f(\lambda)
&=F'(\lambda)\\[9pt]
&=\tfrac2{1-4\lambda^2+\sqrt{1-4\lambda^2}}g\left(\tfrac{2\lambda}{1+\sqrt{1-4\lambda^2}}\right)+\tfrac{1+\sqrt{1-4\lambda^2}}{2\lambda^2\sqrt{1-4\lambda^2}}g\left(\tfrac{1+\sqrt{1-4\lambda^2}}{2\lambda}\right)\tag{3}
\end{align}
$$

Using Only the PDF
Suppose that $Y=g(X)$ and $f_X(x)$ is the PDF for $X$. Consider the case of an infinitesimal interval, $\mathrm{d}x$, in $X$ and the corresponding infinitesimal interval, $\mathrm{d}y$, in $Y$:

When $g'(x)\gt0$, The probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=\frac{f_X(x)}{g'(x)}\,\mathrm{d}y$.
When $g'(x)\lt0$, The probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=-\frac{f_X(x)}{g'(x)}\,\mathrm{d}y$.
Thus, the probability of an event occurring in $\mathrm{d}x$ is $f_X(x)\,\mathrm{d}x=\frac{f_X(x)}{|g'(x)|}\,\mathrm{d}y$.
The probability of and event occurring in $\mathrm{d}y$ is the sum of the probabilities over all the points $x_k$ where $g(x_k)=y$. That is,
$$
f_Y(y)\,\mathrm{d}y=\sum_{g(x_k)=y}\frac{f_X(x_k)}{|g'(x_k)|}\,\mathrm{d}y\tag{4}
$$
Thus, the PDF of $Y$ is
$$
f_Y(y)=\sum_{g(x_k)=y}\frac{f_X(x_k)}{|g'(x_k)|}\tag{5}
$$
Which is the formula cited in Did's answer.
