## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{{x - 3}} - \frac{2}{{x - 1}} + \frac{2}{{x + 1}}$$
\eqalign{ & \frac{{{x^2} - 4x + 11}}{{\left( {x - 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} \cr & {\text{partial fraction decomposition}} \cr & \frac{{{x^2} - 4x + 11}}{{\left( {x - 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} \cr & {x^2} - 4x + 11 = A\left( {x - 1} \right)\left( {x + 1} \right) + B\left( {x - 3} \right)\left( {x + 1} \right) + C\left( {x - 3} \right)\left( {x - 1} \right) \cr & {\text{letting }}x = 3 \cr & 8 = A\left( {3 - 1} \right)\left( {3 + 1} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & 8 = A\left( 2 \right)\left( 4 \right) \cr & 1 = A \cr & {\text{for }}x = 1 \cr & 8 = A\left( 0 \right) + B\left( {1 - 3} \right)\left( {1 + 1} \right) + C\left( 0 \right) \cr & 8 = B\left( { - 2} \right)\left( 2 \right) \cr & - 2 = B \cr & {\text{for }}x = - 1 \cr & 16 = A\left( 0 \right) + B\left( 0 \right) + C\left( { - 4} \right)\left( { - 2} \right) \cr & 16 = C\left( 8 \right) \cr & 2 = C \cr & {\text{substituting the values}} \cr & \frac{A}{{x - 3}} + \frac{B}{{x - 1}} + \frac{C}{{x + 1}} = \frac{1}{{x - 3}} - \frac{2}{{x - 1}} + \frac{2}{{x + 1}} \cr}