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

$[π/6, π/2]$

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

Now, 2 ≤ x² + 2 < ∞  for all x ∈ R

$⇒\frac{1}{2}≥\frac{1}{x^2+2}>0$

$⇒-\frac{1}{2}≤\frac{-1}{x^2+2}<0$

$⇒\frac{1}{2}≤1-\frac{1}{x^2+2}<1$

$⇒\frac{\pi}{6}≤\sin^{-1}(1-\frac{1}{x^2+2})<\frac{\pi}{2}$.

Hence (C) is the correct answer.