## Thomas' Calculus 13th Edition

$$\frac{2}{7}\ln \left| {x + 6} \right| + \frac{5}{7}\ln \left| {x - 1} \right| + C$$
\eqalign{ & \int {\frac{{x + 4}}{{{x^2} + 5x - 6}}} dx \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{{x + 4}}{{{x^2} + 5x - 6}} = \frac{{x + 4}}{{\left( {x + 6} \right)\left( {x - 1} \right)}} \cr & \frac{{x + 4}}{{\left( {x + 6} \right)\left( {x - 1} \right)}} = \frac{A}{{x + 6}} + \frac{B}{{x - 1}} \cr & {\text{Multiplying }}\left( {x + 6} \right)\left( {x - 1} \right){\text{, we have}} \cr & x + 4 = A\left( {x - 1} \right) + B\left( {x + 6} \right) \cr & {\text{if we set }}x = - 6 \cr & - 6 + 4 = A\left( { - 6 - 1} \right) + B\left( 0 \right) \cr & - 2 = A\left( { - 7} \right)\left( 0 \right) \cr & A = \frac{2}{7} \cr & \cr & {\text{if we set }}x = 1 \cr & 1 + 4 = A\left( {1 - 1} \right) + B\left( {1 + 6} \right) \cr & 5 = A\left( 0 \right) + B\left( 7 \right) \cr & B = \frac{5}{7} \cr & \cr & {\text{then}} \cr & \frac{{x + 4}}{{\left( {x + 6} \right)\left( {x - 1} \right)}} = \frac{{2/7}}{{x + 6}} + \frac{{5/7}}{{x - 1}} \cr & \int {\frac{{x + 4}}{{{x^2} + 5x - 6}}} dx = \int {\left( {\frac{{2/7}}{{x + 6}} + \frac{{5/7}}{{x - 1}}} \right)dx} \cr & {\text{integrating}} \cr & = \frac{2}{7}\ln \left| {x + 6} \right| + \frac{5}{7}\ln \left| {x - 1} \right| + C \cr}