## Calculus: Early Transcendentals (2nd Edition)

$$\ln \left| {x - 2} \right| - \frac{4}{{x - 2}} - \frac{2}{{{{\left( {x - 2} \right)}^2}}} + C$$
\eqalign{ & \int {\frac{{{x^2}}}{{{{\left( {x - 2} \right)}^3}}}dx} \cr & {\text{partial fractions}} \cr & \frac{{{x^2}}}{{{{\left( {x - 2} \right)}^3}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{{{\left( {x - 2} \right)}^3}}} \cr & {x^2} = A{\left( {x - 2} \right)^2} + B\left( {x - 2} \right) + C \cr & {\text{multiplying}} \cr & {x^2} = A\left( {{x^2} - 4x + 4} \right) + Bx - 2B + C \cr & {x^2} = A{x^2} - 4Ax + 4A + Bx - 2B + C \cr & {x^2} = \left( {A{x^2}} \right) + \left( { - 4Ax + Bx} \right) + \left( {4A - 2B + C} \right) \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A = 1 \cr & x:{\text{ }} - 4Ax + Bx = 0 \cr & {x^0}:{\text{ }}4A - 2B + C = 0 \cr & {\text{Solving these equations}} \cr & A = 1 \cr & B = 4 \cr & C = 4 \cr & {\text{substituting constants}} \cr & \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} + \frac{C}{{{{\left( {x - 2} \right)}^3}}} = \frac{1}{{x - 2}} + \frac{4}{{{{\left( {x - 2} \right)}^2}}} + \frac{4}{{{{\left( {x - 2} \right)}^3}}} \cr & \int {\frac{{{x^2}}}{{{{\left( {x - 2} \right)}^3}}}dx} = \int {\left( {\frac{1}{{x - 2}} + \frac{4}{{{{\left( {x - 2} \right)}^2}}} + \frac{4}{{{{\left( {x - 2} \right)}^3}}}} \right)dx} \cr & = \int {\frac{1}{{x - 2}}} dx + \int {\frac{4}{{{{\left( {x - 2} \right)}^2}}}} dx + \int {\frac{4}{{{{\left( {x - 2} \right)}^3}}}} dx \cr & {\text{integrating}} \cr & = \ln \left| {x - 2} \right| - \frac{4}{{x - 2}} + \frac{{4{{\left( {x - 2} \right)}^{ - 2}}}}{{ - 2}} + C \cr & {\text{simplify}} \cr & = \ln \left| {x - 2} \right| - \frac{4}{{x - 2}} - \frac{2}{{{{\left( {x - 2} \right)}^2}}} + C \cr}