<s>[INST] What did I do wrong in u-substitution The problem: find the indefinite integral of $x(x-1)^2$
I used u-substitution, $u = x-1, x = u+1, du = dx$.
which gave me $(u+1)u^2$. I distributed and got $u^3 + u^2$, and took the integral to get $[(u^4)/4] + [(u^3)/3]$ replacing $u$ gave me an answer of $[((x-1)^4)/4] + [((x-1)^3)/3] + C$.
The answer sheet solved by distributing before integrating rather than u substitution and got $(x^4)/4 - (2x^3)/3 + (x^2)/2 + C$.
I graphed both integrals thinking they would be equivalent $\pm C$, but they don't appear to be, did I do something wrong?
 [/INST] You did nothing "wrong" with your u-substitution. Your evaluation of the indefinite integral is correct. To see this,
Suggestion: 
Expand the binomials in your answer: Expand $(x-1)^4$ and $(x - 1)^3$ in the numerators, respectively, simplify, and take into account the constant value (absorbed by the constant of integration)...The answers will match, up to the constant of integration.  
$$\dfrac{(x-1)^4}{4} + \dfrac{(x-1)^3}{3} + C \quad = \quad \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + \left(\dfrac{1}{12} + C\right)$$
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