## Calculus: Early Transcendentals (2nd Edition)

$$3x + 14\ln \left| {x - 2} \right| - \ln \left| {x - 1} \right| + C$$
\eqalign{ & \int {\frac{{3{x^2} + 4x - 6}}{{{x^2} - 3x + 2}}} dx \cr & {\text{use long division}} \cr & = \int {\left( {3 + \frac{{13x - 12}}{{{x^2} - 3x + 2}}} \right)} dx \cr & {\text{partial fractions}} \cr & \frac{{13x - 12}}{{{x^2} - 3x + 2}} = \frac{{13x - 12}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} \cr & \frac{{13x - 12}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x - 1}} \cr & 13x - 12 = A\left( {x - 1} \right) + B\left( {x - 2} \right) \cr & x = 2 \to 14 = A \cr & x = 1 \to - 1 = B \cr & \frac{{13x - 12}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{{14}}{{x - 2}} - \frac{1}{{x - 1}} \cr & \int {\left( {3 + \frac{{13x - 12}}{{{x^2} - 3x + 2}}} \right)} dx = \int {\left( {3 + \frac{{14}}{{x - 2}} - \frac{1}{{x - 1}}} \right)} dx \cr & {\text{integrate}} \cr & = 3x + 14\ln \left| {x - 2} \right| - \ln \left| {x - 1} \right| + C \cr}