Answer
$$6\ln \left| {x + 3} \right| - {\text{5ln}}\left| {x - 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x - 39}}{{{x^2} - x - 12}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{x - 39}}{{{x^2} - x - 12}} = \frac{{x - 39}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} \cr
& \frac{{x - 39}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 3}} \cr
& x - 39 = A\left( {x + 3} \right) + B\left( {x - 4} \right) \cr
& {\text{For }}x = 4 \cr
& 4 - 39 = A\left( {4 + 3} \right) \to A = - 5 \cr
& {\text{For }}x = - 3 \cr
& - 3 - 39 = B\left( { - 7} \right) \to B = 6 \cr
& \frac{{x - 39}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} = \frac{{ - 5}}{{x - 4}} + \frac{6}{{x + 3}} \cr
& {\text{Therefore}} \cr
& \int {\frac{{x - 39}}{{{x^2} - x - 12}}} dx = \int {\left( {\frac{{ - 5}}{{x - 4}} + \frac{6}{{x + 3}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }} - {\text{5ln}}\left| {x - 4} \right| + 6\ln \left| {x + 3} \right| + C \cr
& = 6\ln \left| {x + 3} \right| - {\text{5ln}}\left| {x - 4} \right| + C \cr} $$
