Mark M. answered 11/15/19
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
∫x2(x - 6)9dx
Let u = x - 6. Then x = u + 6 and dx = du.
So, the given integral can be rewritten as ∫(u + 6)2u9du = ∫[(u2 + 12u + 36)u9]du
= ∫[u11 + 12u10 + 36u9]du = (1/12)u12 + (12/11)u11 + (18/5)u10 + C
= (1/12)(x - 6)12 + (12/11)(x - 6)11 + (18/5)(x - 6)10 + C