Here is the question
This is my attempt. Is this correct?
Correct.
You could have also used the substitution $\displaystyle u = x+3$:
$\displaystyle \displaystyle \blacktriangleright 3\int\frac{r^2}{r+3}\;{dr} = \int\frac{(u-3)^2}{u}\;{du}$
$\displaystyle \displaystyle = 3\int\frac{u^2-6u+9}{u}\;{du} = 3\int{\frac{9}{u}+u-6\;{du}$
$\displaystyle \displaystyle = 27\ln{u}-\frac{3}{2}u^2-18u+k = 27\ln(r+3)-\frac{3}{2}(r+3)^2-18(r+3)+k$