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?