## Calculus: Early Transcendentals (2nd Edition)

$$- \ln \left( 4 \right)$$
\eqalign{ & \int_{ - 1}^2 {\frac{{5x}}{{{x^2} - x - 6}}dx} \cr & {\text{partial fraction decomposition}} \cr & \frac{{5x}}{{{x^2} - x - 6}} = \frac{{5x}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} \cr & \frac{{5x}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x + 2}} \cr & 5x = A\left( {x + 2} \right) + B\left( {x - 3} \right) \cr & {\text{for }}x = 3 \cr & 15 = A\left( {3 + 2} \right) + B\left( {3 - 3} \right) \cr & 3 = A \cr & \cr & {\text{for }}x = - 2 \cr & - 10 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 3} \right) \cr & - 10 = - 5B \cr & 2 = B \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 3}} + \frac{B}{{x + 2}} = \frac{3}{{x - 3}} + \frac{2}{{x + 2}} \cr & \int_{ - 1}^2 {\frac{{5x}}{{{x^2} - x - 6}}dx} = \int_{ - 1}^2 {\left( {\frac{3}{{x - 3}} + \frac{2}{{x + 2}}} \right)dx} \cr & {\text{integrating}} \cr & = \left[ {3\ln \left| {x - 3} \right| + 2\ln \left| {x + 2} \right|} \right]_{ - 1}^2 \cr & {\text{evaluating limits}} \cr & = \left[ {3\ln \left| {2 - 3} \right| + 2\ln \left| {2 + 2} \right|} \right] - \left[ {3\ln \left| { - 1 - 3} \right| + 2\ln \left| { - 1 + 2} \right|} \right] \cr & = \left[ {3\ln \left( 1 \right) + 2\ln \left( 4 \right)} \right] - \left[ {3\ln \left( 4 \right) + 2\ln \left( 1 \right)} \right] \cr & = 2\ln \left( 4 \right) - 3\ln \left( 4 \right) \cr & = - \ln \left( 4 \right) \cr}