# 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?

• See: wolframalpha.com/input/…. They are equivalent. May 7 '13 at 1:15
• I didn't realize wolfram would show alternate and expanded forms, that's really handy, thanks :D May 7 '13 at 1:18
• incidentally, when I graphed I accidentally divided x^2 by 2 twice, which is why the graphs weren't lining up. May 7 '13 at 1:24
• While using the substitution is a bit excessive for this integrand, keep it in mind for use in clearing out a power of a binomial in something like $\ \int x(x+3)^8 \ dx \ \rightarrow \ \int (u-3) \cdot u^8 \ du \$ ... May 7 '13 at 1:52
• I don't think it was overkill. I did the u sub just as fast if not faster than distributing the square. May 7 '13 at 3:53

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.
$$\dfrac{(x-1)^4}{4} + \dfrac{(x-1)^3}{3} + C \quad = \quad \frac{x^4}{4} - \frac{2x^3}{3} + \frac{x^2}{2} + \left(\dfrac{1}{12} + C\right)$$