Existence of Inverse Mapping : $f: \;P(\Bbb R) \rightarrow \Bbb R $ where $f(x) = {1\over x^2}-x\cdot arctanx+{1 \over 2}log(1+x^2) $ I'd like to show that below function holds the inverse mapping:
$f: \;P(\Bbb R) \rightarrow \Bbb R $ where $f(x) = {1\over x^2}-x\cdot arctanx+{1 \over 2}log(1+x^2) $
To show the existence of inverse mapping, I want to use the property that every inverse mapping has a bijection between domain and range of its orignial function.
But each term of $f(x)$ holds different domain and range which makes this process cubersome.
Any brief approach to check the inversibility of the given function?

Hints (Every monotone function has a inverse mapping):
Since $$f'(x)=-\frac{1}{x^3}-\arctan x,$$
we can conclude that $f'(x) > 0$ whenever $x<0$; $f'(x) < 0$ whenever $x>0$, which leads that $f(x)$ has a inverse mapping.
