Answer
$$\ln \left| x \right| + 2\ln \left| {x + 3} \right| - \ln \left| {x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} - 9x - 9}}{{{x^3} - 9x}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{2{x^2} - 9x - 9}}{{{x^3} - 9x}} = \frac{{2{x^2} - 9x - 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{2{x^2} - 9x - 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{x + 3}} + \frac{C}{{x - 3}} \cr
& {\text{Multiplying the equation by }}x\left( {x + 3} \right)\left( {x - 3} \right){\text{, we have}} \cr
& 2{x^2} - 9x - 9 = A\left( {x + 3} \right)\left( {x - 3} \right) + Bx\left( {x - 3} \right) + Cx\left( {x + 3} \right) \cr
& {\text{if we set }}x = 0 \cr
& 2{\left( 0 \right)^2} - 9\left( 0 \right) - 9 = A\left( 3 \right)\left( { - 3} \right) + B\left( 0 \right) + C\left( 0 \right) \cr
& - 9 = A\left( { - 9} \right) \cr
& A = 1 \cr
& {\text{if we set }}x = - 3 \cr
& 2{\left( { - 3} \right)^2} - 9\left( { - 3} \right) - 9 = A\left( 0 \right) + B\left( { - 3} \right)\left( { - 3 - 3} \right) + C\left( 0 \right) \cr
& 36 = B\left( {18} \right) \cr
& B = 2 \cr
& {\text{if we set }}x = 3 \cr
& 2{\left( 3 \right)^2} - 9\left( 3 \right) - 9 = A\left( 0 \right) + B\left( 0 \right) + C\left( 3 \right)\left( {3 + 3} \right) \cr
& - 18 = C\left( {18} \right) \cr
& C = - 1 \cr
& \cr
& {\text{then}} \cr
& \frac{{2{x^2} - 9x - 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{1}{x} + \frac{2}{{x + 3}} + \frac{{ - 1}}{{x - 3}} \cr
& \cr
& \int {\frac{{2{x^2} - 9x - 9}}{{{x^3} - 9x}}} dx = \int {\left( {\frac{1}{x} + \frac{2}{{x + 3}} + \frac{{ - 1}}{{x - 3}}} \right)dx} \cr
& {\text{integrating}} \cr
& = \ln \left| x \right| + 2\ln \left| {x + 3} \right| - \ln \left| {x - 3} \right| + C \cr} $$
