## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{5}\ln \left| {x - 1} \right| + \frac{2}{5}\ln \left( {{x^2} + 4} \right) + \frac{2}{5}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C$$
\eqalign{ & \int {\frac{{{x^2}}}{{{x^3} - {x^2} + 4x - 4}}dx} \cr & {\text{integrand}} \cr & = \frac{{{x^2}}}{{{x^3} - {x^2} + 4x - 4}} \cr & = \frac{{{x^2}}}{{{x^2}\left( {x - 1} \right) + 4\left( {x - 1} \right)}} \cr & = \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} \cr & {\text{partial fractions}} \cr & \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} \cr & {\text{multiplying}} \cr & {x^2} = A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)\left( {x - 1} \right) \cr & {x^2} = A{x^2} + 4A + B{x^2} - Bx + Cx - C \cr & {x^2} = \left( {A{x^2} + B{x^2}} \right) + \left( { - Bx + Cx} \right) + 4A - C \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A + B = 1 \cr & x:{\text{ }} - B + C = 0 \cr & {x^0}:{\text{ }}4A - C = 0 \cr & {\text{Solving these equations}} \cr & A = 1/5 \cr & B = 4/5 \cr & C = 4/5 \cr & {\text{substituting constants}} \cr & \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} = \frac{{1/5}}{{x - 1}} + \frac{{\left( {4/5} \right)x + \left( {4/5} \right)}}{{{x^2} + 4}} \cr & = \int {\frac{{{x^2}}}{{{x^3} - {x^2} + 4x - 4}}dx} = \int {\left( {\frac{{1/5}}{{x - 1}} + \frac{{\left( {4/5} \right)x + \left( {4/5} \right)}}{{{x^2} + 4}}} \right)dx} \cr & = \int {\frac{{1/5}}{{x - 1}}dx} + \frac{4}{5}\int {\frac{x}{{{x^2} + 4}}dx} + \frac{4}{5}\int {\frac{1}{{{x^2} + 4}}dx} \cr & {\text{integrating}} \cr & = \frac{1}{5}\ln \left| {x - 1} \right| + \frac{2}{5}\ln \left( {{x^2} + 4} \right) + \frac{2}{5}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr}