Answer
$$\frac{{{x^2}}}{2} - 3x + \ln \left| {x + 3} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} - 8}}{{x + 3}}} dx \cr
& {\text{The integrand is an improper rational function since the numerator }} \cr
& {\text{has degree 2 and the denominator 1}}{\text{. Thus}}{\text{, using the long division}} \cr
& \,\,\,\,\,\,\frac{{{x^2} - 8}}{{x + 3}} = x - 3 + \frac{1}{{x + 3}} \cr
& {\text{The integrand can be expressed as}} \cr
& \int {\left( {x - 3 + \frac{1}{{x + 3}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{2} - 3x + \ln \left| {x + 3} \right| + C \cr} $$
