## Calculus: Early Transcendentals (2nd Edition)

$$- \ln \left| x \right| + \frac{9}{x} + \ln \left| {x - 9} \right| + C$$
\eqalign{ & \int {\frac{{81}}{{{x^3} - 9{x^2}}}dx} \cr & {\text{integrand }} \cr & = \frac{{81}}{{{x^3} - 9{x^2}}} \cr & {\text{factor}} \cr & = \frac{{81}}{{{x^2}\left( {x - 9} \right)}} \cr & {\text{partial fractions}} \cr & \frac{{81}}{{{x^2}\left( {x - 9} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 9}} \cr & 81 = Ax\left( {x - 9} \right) + B\left( {x - 9} \right) + C{x^2} \cr & {\text{multiplying}} \cr & 81 = A{x^2} - 9Ax + Bx - 9B + C{x^2} \cr & 81 = \left( {A{x^2} + C{x^2}} \right) + \left( { - 9Ax + Bx} \right) - 9B \cr & {\text{by equating the coefficients}} \cr & {x^2}:{\text{ }}A + C = 0 \cr & x:{\text{ }} - {\text{9}}A + B = 0 \cr & {x^0}:{\text{ }} - 9B = 81 \cr & {\text{Solving these equations}} \cr & B = - 9 \cr & A = - 1 \cr & C = 1 \cr & {\text{substituting constants}} \cr & \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 9}} = \frac{{ - 1}}{x} + \frac{{ - 9}}{{{x^2}}} + \frac{1}{{x - 9}} \cr & \int {\frac{{81}}{{{x^3} - 9{x^2}}}dx} = \int {\left( {\frac{{ - 1}}{x} + \frac{{ - 9}}{{{x^2}}} + \frac{1}{{x - 9}}} \right)dx} \cr & {\text{integrating}} \cr & = - \ln \left| x \right| + \frac{9}{x} + \ln \left| {x - 9} \right| + C \cr}