## Calculus: Early Transcendentals (2nd Edition)

$$\ln \left| {x - 2} \right| - \ln \left| {x + 6} \right| + C$$
\eqalign{ & \int {\frac{8}{{\left( {x - 2} \right)\left( {x + 6} \right)}}} dx \cr & {\text{partial fraction decomposition}} \cr & \frac{8}{{\left( {x - 2} \right)\left( {x + 6} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x + 6}} \cr & 8 = A\left( {x + 6} \right) + B\left( {x - 2} \right) \cr & {\text{for }}x = 2 \cr & 8 = A\left( {2 + 6} \right) + B\left( {2 - 2} \right) \cr & 8 = 8A \cr & 1 = A \cr & {\text{for }}x = - 6 \cr & 8 = A\left( { - 6 + 6} \right) + B\left( { - 6 - 2} \right) \cr & 8 = - 8B \cr & - 1 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 2}} + \frac{B}{{x + 6}} = \frac{1}{{x - 2}} - \frac{1}{{x + 6}} \cr & \int {\frac{8}{{\left( {x - 2} \right)\left( {x + 6} \right)}}} dx = \int {\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 6}}} \right)} dx \cr & {\text{sum rule}} \cr & = \int {\frac{1}{{x - 2}}} dx - \int {\frac{1}{{x + 6}}} dx \cr & {\text{integrating}} \cr & = \ln \left| {x - 2} \right| - \ln \left| {x + 6} \right| + C \cr}