If $f\left(x\right)=\frac{4x(x^2+1)}{x^2+{(x^2+1)}^2},x\ge 0$ then range of f (x) is

$y=\frac{4x(x^2+1)}{x^2+{(x^2+1)}^2}=\frac{4(x+\frac{1}{x})}{1+{(x+\frac{1}{x})}^2}$

Let  $x+\frac{1}{x}=t$
For  $x>0,t\ge 2y=\frac{4t}{1+t^2}=\frac{4}{\frac{1}{t+1}}$
For  $t\ge 2,t+\frac{1}{t}\ge \frac{5}{2}$
$0<y\le \frac{8}{5}$ and y = 0 for x = 0
$\therefore$ Range of f (x) is   $[0,\frac{8}{5}]$