## Calculus: Early Transcendentals (2nd Edition)

$$- \frac{1}{{3\left( {x - 6} \right)}} + \frac{4}{{3\left( {x + 3} \right)}}$$
\eqalign{ & \frac{{x - 9}}{{{x^2} - 3x - 18}} \cr & {\text{factoring}} \cr & = \frac{{x - 9}}{{\left( {x - 6} \right)\left( {x + 3} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{{x - 9}}{{\left( {x - 6} \right)\left( {x + 3} \right)}} = \frac{A}{{x - 6}} + \frac{B}{{x + 3}} \cr & x - 9 = A\left( {x + 3} \right) + B\left( {x - 6} \right) \cr & {\text{letting }}x = 6 \cr & 6 - 9 = A\left( {6 + 3} \right) + B\left( {6 - 6} \right) \cr & - 3 = A\left( 9 \right) \cr & A = - 1/3 \cr & {\text{letting }}x = - 3 \cr & - 3 - 9 = A\left( { - 3 + 3} \right) + B\left( { - 3 - 6} \right) \cr & - 12 = B\left( { - 9} \right) \cr & B = 4/3 \cr & {\text{Substituting the values}} \cr & \frac{A}{{x - 6}} + \frac{B}{{x + 3}} = \frac{{ - 1/3}}{{x - 6}} + \frac{{4/3}}{{x + 3}} \cr & = - \frac{1}{{3\left( {x - 6} \right)}} + \frac{4}{{3\left( {x + 3} \right)}} \cr}