TITLE: How to find the inverse of $f(x)=x+ \frac{x^{3}}{1+x^{2}}$?
QUESTION [10 upvotes]: I know that given,  $f(x)=x+ \frac{x^{3}}{1+x^{2}}$
I should set $y=x+ \frac{x^{3}}{1+x^{2}}$ and solve in terms of $x$, then just swap the $x$'s and $y$'s.
I know that, since the derivative is always positive, and since the function is composed of polynomials, it is continuous and one-to-one, so that an inverse exists. But I cant seem to figure out the algebra to solve in terms of $x$?
I just end up going in circles.
Any help would be greatly appreciated.

REPLY [4 votes]: You get a cubic function/equation for $x$:   
$2x^3 - y \cdot x^2 + x - y = 0$   
The solution for $x$ seems to be as follows:      
$x = - \frac{1}{6} \cdot (-y + C + \frac{y^2-6}{C})$  
where
$C = \sqrt[3]{-y^3-45y + \sqrt{108y^4+1917y^2+216}}$ 
I just followed (by hand) the formulas given here:
general formula for the roots (of a cubic function)