Integration of $ \int x^{2} \sqrt{2x-6} dx $ 
$$ \int x^{2} \sqrt{2x-6} dx = ?$$

My Attempt:
by partial integration
$$ \int x^{2} \sqrt{2x-6} dx = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \int x(2x-6)^{3/2}dx$$
continuing partial integration
$$ = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \left[ \frac{x(2x-6)^{5/2}}{5} - \int \frac{(2x-6)^{5/2}}{5} dx\right]  $$
$$ = (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C$$
Is this the correct and best/simplest answer? 
Strangely, the multiple choices only include answers in the form:
$$ A(2x+6)^{7/2} + B(2x+6)^{5/2} + C(2x+6)^{3/2} + D$$
where $A,B,C,D$ are constants.

Your answer is correct. Compare it with WA answer:
$$= (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C=\\
(2x-6)^{3/2}\left(\frac{x^2}{3}-\frac{2x(2x-6)}{15}+\frac{2(2x-6)^2}{105}\right)+C=\\
(2x-6)^{3/2}\cdot \frac{35x^2-28x^2+84x+8x^2-48x+72}{105}+C=\\
(2x-6)^{3/2}\cdot \frac{5x^2+12x+24}{35}+C=\\
2\sqrt{2}(x-3)^{3/2}\cdot \frac{5x^2+12x+24}{35}+C.$$
I don't think it can be transformed to the expected multiple choices with $2x+6$ as the binomial is under square root. So, I assume there is a typo in the source.
You can avoid the hard methods and calculate as follows:
$$\int x^{2} \sqrt{2x-6} dx = \frac14\int (2x)^{2} \sqrt{2x-6} dx = \frac14\int (2x-6+6)^{2} \sqrt{2x-6} dx = \\
\frac14\int (2x-6)^{2.5}dx+\frac14\int 12(2x-6)^{1.5}dx+\frac14\int 36(2x-6)^{0.5}dx=\\
\frac1{28}(2x-6)^{3.5}+\frac35(2x-6)^{2.5}+3(2x-6)^{1.5}+C.$$
