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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?

The Chaz 2.0
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Daniel B.
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    See: http://www.wolframalpha.com/input/?i=%5B%28%28x-1%29%5E4%29%2F4%5D+%2B+%5B%28%28x-1%29%5E3%29%2F3%5D. They are equivalent. – Ian Coley May 07 '13 at 01:15
  • I didn't realize wolfram would show alternate and expanded forms, that's really handy, thanks :D – Daniel B. May 07 '13 at 01:18
  • incidentally, when I graphed I accidentally divided x^2 by 2 twice, which is why the graphs weren't lining up. – Daniel B. May 07 '13 at 01:24
  • 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 \ $ ... – colormegone May 07 '13 at 01:52
  • I don't think it was overkill. I did the u sub just as fast if not faster than distributing the square. – Daniel B. May 07 '13 at 03:53

1 Answers1

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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.

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

amWhy
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