## Calculus: Early Transcendentals (2nd Edition)

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