The range of $f\left(x\right)=4{\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)$ is

Given,

$f\left(x\right)=4{\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)$

Now taking, $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}<1$

$\Rightarrow 0\le \frac{x^2}{1+x^2}<1$

Now taking ${\sin }^{-1}$ we get,

$\Rightarrow 0\le {\sin }^{-1}\left(\frac{x^2}{1+x^2}\right)<\frac{\pi }{2}$

$\Rightarrow 0\le 4{\sin }^{-1}\left(\frac{x^2}{1+x^2}\right)<2\pi$

Hence, the range of $4{\sin }^{-1}\left(\frac{x^2}{x^2+1}\right)$ is $[0,2\mathrm{\pi })$