1. ## U Substitution

Could someone check if I have the right answer for this? I tried wolfram alpha but it used a long division method instead of u substitution.
$\displaystyle \begin{array}{lll} \int x(x-3)^2 \: dx & = & \int (u+3)u^2 \: du \\ & = & \int u^3 + 3u^2 \: du \\ & = & \frac{u^4}{4} + u^2 + c \\ & = & \frac{(x-3)^4}{4} + (x-3)^2 + c \end{array}$

2. @ alexgeek

You indeed have got the right answer. It is integration by simple substitution. No offences but I wonder why do you even need to check this on the forum ?

3. Presumably, you let u= x- 3 so that du= dx and u+ 3= x. That gives, of course,
$\displaystyle \int x(x- 3)^2dx= \int(u+3)u^2 du$ as you have.

I have no idea what you mean by "a long division method". You could, of course, do this by multiplying: $\displaystyle (x- 3)^2= x^2- 6x+ 9$ so $\displaystyle \int x(x- 3)^2 dx= \int x^3- 6x^2+ 9x dx$.

4. I wanted to check because it looks nothing like the answer wolframalpha produced, so thought it may be wrong.
Cheers.