Inverse of $\frac{x}{1-x^2}$ How did Munkres (page 104 – new international edition) invert the function
$$
  F(x) = \frac{x}{1-x^2} \,, \quad x \in (-1,1)
$$
for
$$
  G(y) = \frac{2y}{1 + \sqrt{1+4y^2}} \,, \quad y \in \mathbb{R} \,?
$$
Looks like some trig-substitution but cannot see it.
 A: The usual method to obtain the inverse is to let $F(x)=y$, interchange $x$ and $y$, and solve for $y$. Thus we solve
$$x=\frac{y}{1-y^2}$$ 
for $y$:
\begin{align*}
x&=\frac{y}{1-y^2}\\
\implies x-xy^2&=y\\
\implies xy^2+y-x&=0\\
\implies y^2+\tfrac1xy-1&=0\\
\implies(y+\tfrac1{2x})^2&=\frac{1+4x^2}{4x^2}\qquad\text{(completing the square)}\\[4pt]
\implies y+\frac{1}{2x} &=\frac{\sqrt{1+4x^2}}{2x}\\
\implies y&=\frac{-1+\sqrt{1+4x^2}}{2x},
\end{align*}
which is equivalent to $G(x)$ if you multiply the top and bottom by $-1-\sqrt{1+4x^2}$. 

Think of this last extra step as similar to realising the denominator when you multiply by the complex conjugate. You can also get to it straight away by using the less well-known Muller's version of the quadratic formula:
$$ x= \frac{-2c}{b\mp\sqrt{b^2-4ac}}.$$
(This version has the advantage that if $a=0$, you still get valid roots).
A: We have $$y=\frac{x}{1-x^2}$$ so we get
$$y-yx^2=x$$ or
$$x^2+\frac{x}{y}-1=0$$ using the quadratic formula we get
$$x_{1,2}=-\frac{1}{2y}\pm \sqrt{\frac{1}{4y^2}+1}$$
A: To invert the function $y=F(x)$, you need to solve for $y$ in the equation $x=F(y)$.  So solve for $y$ in
$$ x = \frac{y}{1-y^2} \implies 0 = y^2 x + y - x \implies y = F^{-1}(x) = \frac{-1 \pm \sqrt{1 + 4x^2}}{2x} $$
where the last equality is quadratic formula
