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Find the integral of x/(x-1)^2 with respect to x

  1. Oct 16, 2004 #1
    Integral (x / (x-1)^2) dx =

    Let u = x, du = dx
    Let dv = 1 / (x-1)^2 , v = -1 / x-1

    uv - integral (vdu) dv

    x ( -1 / (x-1) ) + integral (1 / x-1 ) =

    x ( -1 / (x-1) ) + ln | x - 1 |

    how come the answer is ( -1 / (x-1) ) + ln | x - 1 | ? what happend to the x in ->>> x ( -1 / (x-1) ) + ln | x - 1 | ?

    Thank you in advance
     
  2. jcsd
  3. Oct 16, 2004 #2

    Hurkyl

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    What happened to the constant of integration?
     
  4. Oct 16, 2004 #3
    I'm using this in differential equations I don't need the constant now.
     
  5. Oct 16, 2004 #4

    Hurkyl

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    Ah, but an indefinite integral is a multivalued operation; you always need the constant of integration!

    It turns out that this is problem is a good example of why it matters! When you account for the constant of integration, you can show that your answer and the "right" answer are the same!
     
  6. Oct 16, 2004 #5
    May you show me how I can elliminate the x please?
     
  7. Oct 16, 2004 #6

    Hurkyl

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    Subtract your answer from the "right" answer: what's left?
     
  8. Oct 16, 2004 #7
    thank you SO much!
     
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