Answer
$$\ln \left| {x + \sqrt {{x^2} - 4} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {{x^2} - 4} }}} dx \cr
& {\text{Refer to the triangle below}} \cr
& \sec \theta = \frac{x}{2},{\text{ }}x = 2\sec \theta ,{\text{ }}dx = 2\sec \theta \tan \theta d\theta \cr
& \int {\frac{1}{{\sqrt {{x^2} - 4} }}} dx = \int {\frac{1}{{\sqrt {{{\left( {2\sec \theta } \right)}^2} - 4} }}} \left( {2\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{1}{{\sqrt {4{{\sec }^2}\theta - 4} }}} \left( {2\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{1}{{\sqrt {4\left( {{{\sec }^2}\theta - 1} \right)} }}} \left( {2\sec \theta \tan \theta } \right)d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\frac{1}{{\sqrt {4{{\tan }^2}\theta } }}} \left( {2\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\frac{1}{{2\tan \theta }}} \left( {2\sec \theta \tan \theta } \right)d\theta \cr
& = \int {\sec \theta } d\theta \cr
& {\text{Integrating}} \cr
& = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Write in terms of }}x{\text{, sec}}\theta = \frac{x}{2},{\text{ tan }}\theta = \frac{{\sqrt {{x^2} - 4} }}{2} \cr
& = \ln \left| {\frac{x}{2} + \frac{{\sqrt {{x^2} - 4} }}{2}} \right| + C \cr
& = \ln \left| {\frac{{x + \sqrt {{x^2} - 4} }}{2}} \right| + C \cr
& = \ln \left| {x + \sqrt {{x^2} - 4} } \right| - \ln \left| 2 \right| + C \cr
& {\text{Combine constants}} \cr
& = \ln \left| {x + \sqrt {{x^2} - 4} } \right| + C \cr} $$
