Answer
$3\ln \left| x \right| - 2\ln \left| {x - 4} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{x - 12}}{{{x^2} - 4x}}} dx \cr
& {\text{Factoring the integrand}} \cr
& = \int {\frac{{x - 12}}{{x\left( {x - 4} \right)}}} dx \cr
& {\text{Using the method of partial fractions}} \cr
& \frac{{x - 12}}{{x\left( {x - 4} \right)}} = \frac{A}{x} + \frac{B}{{x - 4}} \cr
& {\text{Multiply the equation by }}x\left( {x - 4} \right) \cr
& x - 12 = A\left( {x - 4} \right) + Bx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Let }}x = 0{\text{ into }}\left( {\bf{1}} \right) \cr
& 0 - 12 = A\left( {0 - 4} \right) + B\left( 0 \right){\text{ }} \to A = 3 \cr
& {\text{Let }}x = 4{\text{ into }}\left( {\bf{1}} \right) \cr
& 4 - 12 = A\left( {4 - 4} \right) + B\left( 4 \right){\text{ }} \to B = - 2 \cr
& {\text{Substituting }}A{\text{ and }}B{\text{ into the partial fraction decomposition}} \cr
& \frac{{x - 12}}{{x\left( {x - 4} \right)}} = \frac{3}{x} - \frac{2}{{x - 4}} \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{x - 12}}{{{x^2} - 4x}}} dx = \int {\left( {\frac{3}{x} - \frac{2}{{x - 4}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = 3\ln \left| x \right| - 2\ln \left| {x - 4} \right| + C \cr} $$