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How did the book do this integration

  1. Oct 18, 2004 #1
    I'm sure it's pretty simple, but I'm just not seeing it.

    Integrate (x^2/(x-1)) dx

    I did it with a u substitution, letting u = x-1 and then x = u+1

    which ultimately leads me to integrate (u^2/u) + (2u/u) + (1/u)

    After canceling, integrating, and substituting I'm left with
    (x^2/2) + x + ln(abs)(x-1) + C (I assume I'm alright rolling the -3/2 that I had left over into C to make it match the book answer?)

    The book does it like this and I'm not sure what's going on
    Integrate (x^2/(x-1)) dx = Integrate (x+1) dx + Integrate (1/(x-1)) dx

    I think they are splitting it up somehow, hence the 1/x-1, but I'm not sure how they got to x+1

    which yields the same answer I had.
  2. jcsd
  3. Oct 18, 2004 #2


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  4. Oct 18, 2004 #3


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    All they are doing is rewriting the integrand this way:

    [tex]\frac {x^2}{x-1} = \frac {x^2 -1 + 1}{x-1} = \frac {x^2-1}{x-1}+\frac {1}{x-1} = x+1 + \frac {1}{x-1}[/tex]
  5. Oct 18, 2004 #4
    Bah, I must have been way off my game this morning, thanks.
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