Answer
$$\ln \left| {x + \sqrt {{x^2} - 4} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {{x^2} - 4} }}} \cr
& {\text{substitute }}u = x,{\text{ }}du = dx \cr
& = \int {\frac{{dx}}{{\sqrt {{x^2} - 4} }} = } \int {\frac{{dx}}{{\sqrt {{x^2} - {{\left( 2 \right)}^2}} }}} \cr
& = \int {\frac{{du}}{{\sqrt {{u^2} - {2^2}} }}} \cr
& {\text{find the antiderivative }}\left( {{\text{formula 25 page 490}}} \right) \cr
& = \ln \left| {u + \sqrt {{u^2} - {{\left( 2 \right)}^2}} } \right| + C \cr
& = \ln \left| {u + \sqrt {{u^2} - 4} } \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = x \cr
& = \ln \left| {x + \sqrt {{x^2} - 4} } \right| + C \cr} $$
