Find the range of tan^-1(2x/(1+x^2))

First, we must get the range of

$\frac{2x}{1+x^2}=y$

We have $yx^2 - 2x +y = 0$

Since x is real, D ≥ 0, i.e., $4 - 4y^2 ≥0$ or $-1≤ y ≤1$.

So, $\tan^{-1}(y)∈[-\frac{π}{4},\frac{π}{4}]$  (As tan x is an increasing function)