Answer
$$9x + 2\ln \left| x \right| + \frac{1}{x} + 7\ln \left| {x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{9{x^3} - 3x + 1}}{{{x^3} - {x^2}}}} dx \cr
& \cr
& {\text{Apply long division to the integrand}} \cr
& \frac{{9{x^3} - 3x + 1}}{{{x^3} - {x^2}}} = 9 + \frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{9{x^3} - 3x + 1}}{{{x^3} - {x^2}}}} dx = \int {\left( {9 + \frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}}} \right)} dx \cr
& = \int 9 dx + \int {\frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}}} dx \cr
& = 9x + \int {\frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}}} dx \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}} = \frac{{9{x^2} - 3x + 1}}{{{x^2}\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 1}} \cr
& {\text{Multiplying by }}{x^2}\left( {x - 1} \right){\text{, we have}} \cr
& 9{x^2} - 3x + 1 = Ax\left( {x - 1} \right) + B\left( {x - 1} \right) + C{x^2} \cr
& \cr
& {\text{if we set }}x = 0, \cr
& 1 = A\left( 0 \right) + B\left( { - 1} \right) + C\left( 0 \right) \cr
& B = - 1 \cr
& {\text{if we set }}x = 1 \cr
& 7 = A\left( 0 \right) + B\left( 0 \right) + C{\left( 1 \right)^2} \cr
& C = 7 \cr
& {\text{if we set }}x = 2,\,\,B = - 1{\text{ and }}C = 7 \cr
& 9{\left( 2 \right)^2} - 3\left( 2 \right) + 1 = A\left( 2 \right)\left( {2 - 1} \right) + \left( { - 1} \right)\left( {2 - 1} \right) + \left( 7 \right){\left( 2 \right)^2} \cr
& 31 = 2A - 1 + 28 \cr
& A = 2 \cr
& \frac{{9{x^2} - 3x + 1}}{{{x^2}\left( {x - 1} \right)}} = \frac{2}{x} - \frac{1}{{{x^2}}} + \frac{7}{{x - 1}} \cr
& \int {\frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}}} dx = \int {\left( {\frac{2}{x} - \frac{1}{{{x^2}}} + \frac{7}{{x - 1}}} \right)} dx \cr
& = 2\ln \left| x \right| + \frac{1}{x} + 7\ln \left| {x - 1} \right| + C \cr
& {\text{Then}}{\text{,}} \cr
& 9x + \int {\frac{{9{x^2} - 3x + 1}}{{{x^3} - {x^2}}}} dx = 9x + 2\ln \left| x \right| + \frac{1}{x} + 7\ln \left| {x - 1} \right| + C \cr} $$
