Answer
$$x + - \frac{{64}}{{11}}\ln \left| {x + 8} \right| + \frac{9}{{11}}\ln \left| {x - 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{x^2} + 5x - 24}}} dx \cr
& {\text{By long division}} \cr
& \frac{{{x^2}}}{{{x^2} + 5x - 24}} = 1 + \frac{{ - 5x + 24}}{{{x^2} + 5x - 24}} \cr
& = \int {\left( {1 + \frac{{ - 5x + 24}}{{{x^2} + 5x - 24}}} \right)} dx \cr
& = x + \int {\frac{{ - 5x + 24}}{{{x^2} + 5x - 24}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Decompose }}\frac{{ - 5x + 24}}{{{x^2} + 5x - 24}}{\text{ into partial fractions}} \cr
& \frac{{ - 5x + 24}}{{{x^2} + 5x - 24}} = \frac{{ - 5x + 24}}{{\left( {x + 8} \right)\left( {x - 3} \right)}} \cr
& \frac{{ - 5x + 24}}{{\left( {x + 8} \right)\left( {x - 3} \right)}} = \frac{A}{{x + 8}} + \frac{B}{{x - 3}} \cr
& - 5x + 24 = A\left( {x - 3} \right) + B\left( {x + 8} \right) \cr
& x = - 8 \to - 5\left( { - 8} \right) + 24 = A\left( { - 8 - 3} \right) \to A = - \frac{{64}}{{11}} \cr
& x = 3 \to - 5\left( 3 \right) + 24 = B\left( {3 + 8} \right) \to B = \frac{9}{{11}} \cr
& {\text{Then,}} \cr
& \frac{{ - 5x + 24}}{{{x^2} + 5x - 24}} = \frac{{ - \frac{{64}}{{11}}}}{{x + 8}} + \frac{{\frac{9}{{11}}}}{{x - 3}} \cr
& \int {\frac{{ - 5x + 24}}{{{x^2} + 5x - 24}}} dx = \int {\left( {\frac{{ - \frac{{64}}{{11}}}}{{x + 8}} + \frac{{\frac{9}{{11}}}}{{x - 3}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }} - \frac{{64}}{{11}}\ln \left| {x + 8} \right| + \frac{9}{{11}}\ln \left| {x - 3} \right| + C \cr
& {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr
& = x + - \frac{{64}}{{11}}\ln \left| {x + 8} \right| + \frac{9}{{11}}\ln \left| {x - 3} \right| + C \cr} $$
