Answer
${x^2} + \frac{3}{2}\ln \left| {x - 4} \right| - \frac{1}{2}\ln \left| {x + 2} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^3} - 4{x^2} - 15x + 5}}{{{x^2} - 2x - 8}}} dx \cr
& {\text{Using the long division}} \cr
& \frac{{2{x^3} - 4{x^2} - 15x + 5}}{{{x^2} - 2x - 8}} = 2x + \frac{{x + 5}}{{{x^2} - 2x - 8}} \cr
& \cr
& = \int {\left( {2x + \frac{{x + 5}}{{{x^2} - 2x - 8}}} \right)} dx \cr
& = \int {2x} dx + \int {\frac{{x + 5}}{{{x^2} - 2x - 8}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Decomposing }}\frac{{x + 5}}{{{x^2} - 2x - 8}}{\text{ into partial fractions}} \cr
& \frac{{x + 5}}{{\left( {x - 4} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 2}} \cr
& x + 5 = A\left( {x + 2} \right) + B\left( {x - 4} \right) \cr
& {\text{Let }}x = 4 \cr
& 9 = A\left( 6 \right) + B\left( 0 \right) \to A = \frac{3}{2} \cr
& {\text{Let }}x = - 2 \cr
& 3 = A\left( 0 \right) + B\left( { - 6} \right) \to B = - \frac{1}{2} \cr
& {\text{Therefore}} \cr
& \frac{{x + 5}}{{\left( {x - 4} \right)\left( {x + 2} \right)}} = \frac{{3/2}}{{x - 4}} + \frac{{ - 1/2}}{{x + 2}} \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& = \int {2x} dx + \int {\left( {\frac{{3/2}}{{x - 4}} + \frac{{ - 1/2}}{{x + 2}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = {x^2} + \frac{3}{2}\ln \left| {x - 4} \right| - \frac{1}{2}\ln \left| {x + 2} \right| + C \cr} $$
