Q:

Table of Integrals. Which one to use?

I just finished my exam and there was this question. It asks us for which form in the table of integrals to use.
$\int(x-3)\sqrt(6+6x-x^2)$
I did completing the square and got into this form
$\int(x-3)\sqrt(15-(x-3)^2)$
then u-sub with u= x-3
and I got this form
$\int u\sqrt(15-u^2)$
I know I can solve it using substitution again but the question is a MCQ question. It asks us to choose which one to use from the table of integration.
I think the choices were 29,30,11,12. But neither seems to fit my equation?
http://mgh-images.s3.amazonaws.com/9780073532325/1702-6.5-28EI2.png

A:

Hint: If $$\int (x-3)\sqrt{6+6x-x^2}\, dx $$
take $u=6+6x-x^2\implies du=(6-2x)dx=-2(x-3)dx$
then
$$\int (x-3)\sqrt{6+6x-x^2}\, dx=\frac{-1}{2}\int u^{1/2} du $$

