Answer
$$\frac{{{x^2}}}{2} - 2x + 6\ln \left| {x + 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 2}}{{x + 2}}} dx \cr
& {\text{Using long division }}\frac{{{x^2} + 2}}{{x + 2}} = x - 2 + \frac{6}{{x + 2}} \cr
& {\text{Then}}{\text{, the integrand is an improper rational function that can be written as}} \cr
& \int {\frac{{{x^2} + 2}}{{x + 2}}} dx = \int {\left( {x - 2 + \frac{6}{{x + 2}}} \right)} dx \cr
& = \int x dx - 2dx + \int {\frac{6}{{x + 2}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{2} - 2x + 6\ln \left| {x + 2} \right| + C \cr} $$
