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Homework Help: Integration math question

  1. May 26, 2008 #1
    If we were to integrate 1/x by parts using u = 1/x and dv = 1dx, then it would end up as:
    Int(dx/x) = 1 + Int(dx/x), ending up with 1 = 0. Why was there a discrepancy in the integration?
    --Int() refers to integral
     
  2. jcsd
  3. May 26, 2008 #2

    morphism

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    You're missing an integration constant.
     
  4. May 26, 2008 #3

    exk

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    [tex]\int\frac{1}{x}dx=uv -\int vdu= \frac{x+c}{x} - \int \frac{-(x+c)}{x^{2}}dx = etc...[/tex]
    [tex]u=\frac{1}{x}[/tex]
    [tex]du=\frac{-1}{x^{2}}[/tex]
    [tex]dv=1dx[/tex]
    [tex]v=x+c[/tex]

    Please tell me if I am wrong.

    Regardless, you are looking for the result to be ln|x|+c.
     
  5. May 26, 2008 #4

    Defennder

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    I don't think there should be a '+ c' in this intermediate step. Anyway, integrating
    1/x gives ln(x) + c by definition. Some books actually define ln(x) to be the anti-derivative of that.
     
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