## Calculus: Early Transcendentals (2nd Edition)

$$6\ln \left| {x - 6} \right| + 4\ln \left| {x + 4} \right| + C$$
\eqalign{ & \int {\frac{{10x}}{{{x^2} - 2x - 24}}dx} \cr & {\text{partial fraction decomposition}} \cr & \frac{{10x}}{{{x^2} - 2x - 24}} = \frac{{10x}}{{\left( {x - 6} \right)\left( {x + 4} \right)}} \cr & \frac{{10x}}{{\left( {x - 6} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 6}} + \frac{B}{{x + 4}} \cr & 10x = A\left( {x + 4} \right) + B\left( {x - 6} \right) \cr & {\text{for }}x = 6 \cr & 10\left( 6 \right) = A\left( {6 + 4} \right) + B\left( {6 - 6} \right) \cr & 6 = A \cr & \cr & {\text{for }}x = - 4 \cr & 10\left( { - 4} \right) = A\left( { - 4 + 4} \right) + B\left( { - 4 - 6} \right) \cr & - 40 = B\left( { - 4 - 6} \right) \cr & 4 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 6}} + \frac{B}{{x + 4}} = \frac{6}{{x - 6}} + \frac{4}{{x + 4}} \cr & \int {\frac{{10x}}{{{x^2} - 2x - 24}}dx} = \int {\left( {\frac{6}{{x - 6}} + \frac{4}{{x + 4}}} \right)} dx \cr & {\text{sum rule}} \cr & = \int {\frac{6}{{x - 6}}} dx + \int {\frac{4}{{x + 4}}} dx \cr & {\text{integrating}} \cr & = 6\ln \left| {x - 6} \right| + 4\ln \left| {x + 4} \right| + C \cr}