Answer
$$\frac{1}{3}{\left( {{x^2} - 4} \right)^{3/2}} + C $$
Work Step by Step
$$\eqalign{
& \int x \sqrt {{x^2} - 4} dx \cr
& {\text{We set}}{\text{, }}x = 2\sec \theta,{\text{ }}dx = 2\sec \theta \tan \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int x \sqrt {{x^2} - 4} dx = \int {2\sec \theta } \sqrt {{{\left( {2\sec \theta } \right)}^2} - 4} \left( {2\sec \theta \tan \theta d\theta } \right) \cr
& = \int {4{{\sec }^2}\theta \tan \theta } \sqrt {4{{\sec }^2}\theta - 4} d\theta \cr
& = \int {4{{\sec }^2}\theta \tan \theta } \sqrt {4\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {4{{\sec }^2}\theta \tan \theta } \sqrt {4{{\tan }^2}\theta } d\theta \cr
& {\text{simplifying}} \cr
& = \int {8{{\sec }^2}\theta {{\tan }^2}\theta } d\theta \cr
& {\text{integrate}} \cr
& = 8\left( {\frac{{{{\tan }^3}\theta }}{3}} \right) + C \cr
& = \frac{8}{3}{\tan ^3}\theta + C \cr
& {\text{write in terms of }}x,{\text{ tan}}\theta = \frac{{\sqrt {{x^2} - 4} }}{2} \cr
& = \frac{8}{3}{\left( {\frac{{\sqrt {{x^2} - 4} }}{2}} \right)^3} + C \cr
& = \frac{1}{3}{\left( {{x^2} - 4} \right)^{3/2}} + C \cr} $$
