Answer
$$9\ln \left| {x - 4} \right| - 7\ln \left| {x - 3} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{2x + 1}}{{{x^2} - 7x + 12}}} dx \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{2x + 1}}{{{x^2} - 7x + 12}} = \frac{{2x + 1}}{{\left( {x - 4} \right)\left( {x - 3} \right)}} \cr
& \frac{{2x + 1}}{{\left( {x - 4} \right)\left( {x - 3} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x - 3}} \cr
& {\text{Multiplying }}\left( {x - 4} \right)\left( {x - 3} \right){\text{, we have}} \cr
& 2x + 1 = A\left( {x - 3} \right) + B\left( {x - 4} \right) \cr
& \cr
& {\text{if we set }}x = 4 \cr
& 2\left( 4 \right) + 1 = A\left( {4 - 3} \right) + B\left( {4 - 4} \right) \cr
& 9 = A\left( 1 \right) \cr
& A = 9 \cr
& \cr
& {\text{if we set }}x = 3 \cr
& 2\left( 3 \right) + 1 = A\left( {3 - 3} \right) + B\left( {3 - 4} \right) \cr
& 7 = A\left( 0 \right) + B\left( { - 1} \right) \cr
& B = - 7 \cr
& \cr
& {\text{then}} \cr
& \frac{{2x + 1}}{{\left( {x - 4} \right)\left( {x - 3} \right)}} = \frac{9}{{x - 4}} + \frac{{ - 7}}{{x - 3}} \cr
& \int {\frac{{2x + 1}}{{{x^2} - 7x + 12}}} dx = \int {\left( {\frac{9}{{x - 4}} - \frac{7}{{x - 3}}} \right)dx} \cr
& {\text{integrating}} \cr
& = 9\ln \left| {x - 4} \right| - 7\ln \left| {x - 3} \right| + C \cr} $$
