Range of f(x)=sin^-1(√(x^2+x+1)) is

$0 ≤ x^2 + 2×\frac{1}{2}x + 1 ≤ 1$

$⇒\frac{3}{4}≤(x+\frac{1}{2})^2+\frac{3}{4}≤ 1$

so $\frac{\sqrt 3}{2}≤\sqrt{x^2+x+1}≤ 1$

so $\frac{π}{3}≤\sin^{-1}\sqrt{x^2+x+1}≤\frac{π}{2}$