Help integrating a value! :O int((3x-3)^2)

Hi thanks for feeling interest and clicking on the topic. well so. i have integrated the value after how i believe it should be done. my friend said he could verify if it was correct with his programs. but the “microsoft math” he had, and “maple 12” program he had, showed 2 different answers, which were:
micro.: 3x^3-9x^2+9x
maple: 3x^3-9x^2+9x-3
I myself got maples answer, but have always been able to use microsoft math in terms of other standard calculations and are therefore a bit worried which calculated wrong >.<
what i did myself was: u=3x-3
du/dx=g'(x)=3
dt=3dx
dx=1/3 dt
now i can substitute with u:
int((3x-3)^2 dx) <=> int(u^2*1/3du)
the rule of the constant then:
1/3int(u^2 dt)
then the powers rule..:
(1/9)u^3 we revert u back..
(1/9)*(3x-3)^3
(1/9)*(27x^3-81x^2+81x-27)
3x^3-9x^2+9x-3
the same as maple :O but microsoft math has never failed me before ow :X.. (never used it for integrals thou!)
thanks :O
btw: http://integrals.wolfram.com/index.jsp?expr=(3x-3)^2&random=false tried inputting the value which needs to be integrated there too, and it gave me the same answer as microsoft math too. but ofc it can be wrong too, coz something needs to be wrong and i dont know where i have gone wrong :X
Answer #1
Requesting a master of math
Answer #2
any >.<?
Answer #3
must be some geniuses here <.<
Answer #4
not..
Answer #5
sry for spamming this thread like this its an exam hand in for Wednesday