Answer
$$2\left( {x - 2\ln \left| {x + 2} \right|} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} - 4x}}{{{x^2} - 4}}} dx \cr
& {\text{By the long division }}\frac{{2{x^2} - 4x}}{{{x^2} - 4}} = 2 + \frac{{8 - 4x}}{{{x^2} - 4}} \cr
& \int {\frac{{2{x^2} - 4x}}{{{x^2} - 4}}} dx = \int {\left( {2 + \frac{{8 - 4x}}{{{x^2} - 4}}} \right)} dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int 2 dx + \int {\frac{8}{{{x^2} - 4}}} dx - 2\int {\frac{{2x}}{{{x^2} - 4}}} \cr
& {\text{Integrate use }}\int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x + \frac{8}{{2\left( 2 \right)}}\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| - \ln \left| {{x^2} - 4} \right| + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x + 2\ln \left| {\frac{{x - 2}}{{x + 2}}} \right| - \ln \left| {\left( {x + 2} \right)\left( {x - 2} \right)} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x + 2\ln \left( {\left| {\frac{{x - 2}}{{x + 2}} \cdot \frac{1}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right|} \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x + 2\ln \left( {\frac{1}{{{{\left( {x + 2} \right)}^2}}}} \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2x - 2\ln {\left( {x + 2} \right)^2} + C \cr
& {\text{Factoring}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {x - 2\ln \left| {x + 2} \right|} \right) + C \cr} $$
