Answer
$$x\ln \sqrt {{x^2} - 4} - x - \ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \sqrt {{x^2} - 4} } dx \cr
& {\text{Write }}\sqrt {{x^2} - 4} = {\left( {{x^2} - 4} \right)^{1/2}}{\text{ and apply logartihmic properties}} \cr
& = \int {\ln {{\left( {{x^2} - 4} \right)}^{1/2}}} dx \cr
& = \int {\frac{1}{2}\ln \left( {{x^2} - 4} \right)} dx \cr
& = \frac{1}{2}\int {\ln \left( {{x^2} - 4} \right)} dx \cr
& {\text{Integrating by parts}} \cr
& u = \ln \left( {{x^2} - 4} \right),{\text{ }}du = \frac{{2x}}{{{x^2} - 4}}dx \cr
& dv = dx,{\text{ }}v = x \cr
& \int {udv} = uv - \int {vdu} \cr
& \frac{1}{2}\int {\ln \left( {{x^2} - 4} \right)} dx = \frac{1}{2}\left( {x\ln \left( {{x^2} - 4} \right) - \int {x\left( {\frac{{2x}}{{{x^2} - 4}}} \right)dx} } \right) \cr
& = \frac{1}{2}x\ln \left( {{x^2} - 4} \right) - \frac{1}{2}\int {x\left( {\frac{{2x}}{{{x^2} - 4}}} \right)dx} \cr
& = \frac{1}{2}x\ln \left( {{x^2} - 4} \right) - \int {\frac{{{x^2}}}{{{x^2} - 4}}dx} \cr
& = \frac{1}{2}x\ln \left( {{x^2} - 4} \right) - \int {\frac{{{x^2} - 4 + 4}}{{{x^2} - 4}}dx} \cr
& = \frac{1}{2}x\ln \left( {{x^2} - 4} \right) - \int {\left( {1 + \frac{4}{{{x^2} - 4}}} \right)dx} \cr
& = \frac{1}{2}x\ln \left( {{x^2} - 4} \right) - \int {dx} - 4\int {\frac{4}{{{x^2} - 4}}dx} \cr
& {\text{By tables }}\int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C,{\text{ then}}} \cr
& = x\ln {\left( {{x^2} - 4} \right)^{1/2}} - x - 4\left( {\frac{1}{4}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right|} \right) + C \cr
& = x\ln \sqrt {{x^2} - 4} - x - \ln \left| {\frac{{x - 2}}{{x + 2}}} \right| + C \cr} $$
