Answer
$$ = {\cos ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {{x^2} - 2} }}} \cr
& = \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} \cr
& {\text{substitute }}u = x,{\text{ }}du = dx,{\text{ }}a = \sqrt 2 \cr
& = \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( {\sqrt 2 } \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {{u^2} - {a^2}} }}} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& = {\cos ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& {\text{write in terms of }}x \cr
& = {\cos ^{ - 1}}\left( {\frac{x}{{\sqrt 2 }}} \right) + C \cr} $$
