Answer
$$\frac{{{x^2}}}{2} - 4x + 16\ln \left| {x - 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} - 4}}{{x + 4}}} dx \cr
& {\text{long division}} \cr
& = \int {\left( {x - 4 + \frac{{12}}{{x + 4}}} \right)} dx \cr
& = \int x dx - \int {4dx} + \int {\frac{{12}}{{x + 4}}} dx \cr
& {\text{find antiderivatives}} \cr
& = \frac{{{x^2}}}{2} - 4x + 16\ln \left| {x - 4} \right| + C \cr} $$
