<s>[INST] 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
 [/INST] Substitute $u=x+3 \implies x=u-3, dx=du$:$$\int x\sqrt{(x+3)^2-6}dx=\int (u-3)\sqrt{u^2-6} du$$Integrating the integral gives:
$$\int u\sqrt{u^2-6}du-\int 3\sqrt{u^2-6} du=\frac{(u^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{u\sqrt{u^2-6}}{2}-3\ln\left(u+\sqrt{u^2-6}\right)\right)$$In terms of $x$:$$\int x\sqrt{(x+3)^2-6}dx=$$$$\frac{((x+3)^2-6)^{\frac{3}{2}}}{3}-3\left(\frac{(x+3)\sqrt{(x+3)^2-6}}{2}-3\ln\left(x+3+\sqrt{(x+3)^2-6}\right)\right)$$
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