<s>[INST] 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.
 [/INST] Hint:Substituting $$t=\sqrt{2x-6}$$ then we get $$x=\frac{t^2+6}{2}$$ then we get
$$dx=tdt$$
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