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?

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.
