Integration by completing the square I need to complete the square on the following integral.  Once this is done apparently I will be able to use on of the integration tables in the back of my book.
$\int x  \sqrt{x^2 + 6x +3} dx $
This is what I have come up with so far:
$\int x \sqrt{(x+3)^2 -6}  $
I really am at a loss.  
Any help with this would be appreciated.
Thank you

$$ I= \int x \sqrt {x^2 +6x +3} dx $$
This is best done by substitution.
$$I= \int x \sqrt {(x+3)^2-6} dx \\ x+3= \sqrt 6 \sec \theta \implies dx=\sqrt 6 \sec \theta \tan \theta d \theta$$
$$ \therefore I= \sqrt 6 \int \sec \theta \tan \theta ( \sqrt 6 \sec \theta -3) \sqrt {6( \sec^2 \theta -1)}  \\ I=6 \int \sec \theta \tan^2 \theta ( \sqrt 6 \sec \theta -3)d \theta \\ I=6 \sqrt6 \int \sec^2 \theta \tan^2 \theta d\theta-18\int \sec  \theta \tan^2 \theta d \theta \\ \therefore I= 2 \sqrt6 \tan^3 \theta - 18\int \sec^3 \theta d \theta + 18\int \sec \theta d \theta \\ I =2 \sqrt6 \tan^3 \theta -9 \sec \theta \tan \theta +9 \ln | \sec \theta + \tan \theta | +C$$
Unless I screwed up the algebra I think this is correct. It could also be done using the substitution $ u=x^2+6x+3 $ and evaluating $\int \sqrt {x^2+6x+3} dx$.
