# How to find the inverse of $f(x)=x+ \frac{x^{3}}{1+x^{2}}$?

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.

• The answer seems a bit complicated, so maybe you're just best off pointing out that an inverse exists and calling it a day. – Arthur May 31 '16 at 9:51
• @Arthur It's not so complicated actually, takes about 20 mins by hand. All that is needed is a few big sheets of paper :) – peter.petrov May 31 '16 at 10:18

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)