If $\int\limits_1^x\frac{dt}{|t|\sqrt{t^2-1}}=\frac{π}{6}$, then x can be equal to:

$2/\sqrt{3}$ $\sqrt{3}$ 2 None of these

$2/\sqrt{3}$

$\int\limits_1^x\frac{dt}{|t|\sqrt{t^2-1}}=\frac{π}{6}⇒[sec^{-1}(t)]_1^x=\frac{π}{6}⇒sec^{-1}(x)=\frac{π}{6}$ $⇒sec^{-1}(x)=\frac{π}{6}⇒x=\frac{2}{\sqrt{3}}$