Chapter 7 - Principles Of Integral Evaluation - 7.5 Integrating Rational Functions By Partial Fractions - Exercises Set 7.5 - Page 521: 17

$$3x + 12\ln \left| {x - 2} \right| - \frac{2}{{x - 2}} + C$$

$$\eqalign{ & \int {\frac{{3{x^2} - 10}}{{{x^2} - 4x + 4}}} dx \cr & {\text{The integrand is an improper rational function since the numerator }} \cr & {\text{has degree 2 and the denominator 2}}{\text{. Thus}}{\text{, using the long division}} \cr & \,\,\,\,\,\,\frac{{3{x^2} - 10}}{{{x^2} - 4x + 4}} = 3 + \frac{{12x - 22}}{{{x^2} - 4x + 4}} \cr & {\text{The integrand can be expressed as}} \cr & \int {\left( {3 + \frac{{12x - 22}}{{{x^2} - 4x + 4}}} \right)} dx \cr & = 3x + \int {\frac{{12x - 22}}{{{x^2} - 4x + 4}}} dx \cr & {\text{Decomposing the integrand into partial fractions}} \cr & {\text{Factor the denominator}} \cr & \frac{{12x - 22}}{{{x^2} - 4x + 4}} = \frac{{12x - 22}}{{{{\left( {x - 2} \right)}^2}}} \cr & \frac{{12x - 22}}{{{{\left( {x - 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} \cr & {\text{Multiplying the equation by }}{\left( {x - 2} \right)^2}{\text{, we have}} \cr & 12x - 22 = A\left( {x - 2} \right) + B \cr & {\text{if we set }}x = 2 \cr & 12\left( 2 \right) - 22 = A\left( 0 \right) + B \cr & B = 2 \cr & {\text{Then}}{\text{,}}\,\,{\text{set }}x = 0{\text{ and }}B = 2 \cr & 12\left( 0 \right) - 22 = A\left( {0 - 2} \right) + 2 \cr & - 24 = A\left( { - 2} \right) \cr & A = 12 \cr & \cr & {\text{then}} \cr & 3x + \int {\frac{{12x - 22}}{{{x^2} - 4x + 4}}} dx = 3x + \int {\left( {\frac{{12}}{{x - 2}} + \frac{2}{{{{\left( {x - 2} \right)}^2}}}} \right)} dx \cr & = 3x + \int {\frac{{12}}{{x - 2}}} dx + 2\int {{{\left( {x - 2} \right)}^{ - 2}}} dx \cr & {\text{Integrating}} \cr & = 3x + 12\ln \left| {x - 2} \right| + 2\left( {\frac{{{{\left( {x - 2} \right)}^{ - 1}}}}{{ - 1}}} \right) + C \cr & = 3x + 12\ln \left| {x - 2} \right| - \frac{2}{{x - 2}} + C \cr} $$