Range of $\sin^{-1}\left(\frac{x^2+1}{x^2+2}\right)$ is

$\left[\frac{π}{6},\frac{π}{2}\right)$

Here, $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$

Now, $2≤x^2+2<∞$ for all x ∈ R ⇒ $\frac{1}{2}≥\frac{1}{x^2+2}>0⇒-\frac{1}{2}≤\frac{-1}{x^2+2}<0$

$⇒\frac{1}{2}≤1-\frac{1}{x^2+2}<1⇒\frac{π}{6}≤\sin^{-1}\left(1-\frac{1}{x^2+2}\right)<\frac{π}{2}$