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Proof rational integral

  1. Feb 18, 2010 #1
    Dear Forum,
    I've been trying to find a proof for the following:

    [tex]

    \int \frac{x}{a^2+x^2}dx = \frac{1}{2}\ln|a^2+x^2|+c

    [/tex]

    After many hours I've resorted to asking for help - any ideas anyone?

    cheers,
    mazzo
     
  2. jcsd
  3. Feb 18, 2010 #2

    CompuChip

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    Science Advisor
    Homework Helper

    If you just want to check this, you can simply differentiate the right hand side and show that you get the integrand back.

    If you want to calculate it (forgetting about the result for the time being), then you may notice that x is the derivative of 1/2(a2 + x2) and try a variable substitution u = a2 + x2.
     
  4. Feb 18, 2010 #3
    thanks for this. But I found that when I substituted in this value for u it doesn't work. I tried
    [tex]
    x^2 = t
    [/tex]

    [tex]
    2xdx = dt
    [/tex]

    [tex]
    xdx = dt/2
    [/tex]

    then

    [tex]
    \int \frac{x}{a^2+x^2}dx = \frac{1}{2} \int \frac{dt}{(a^2+t)} = \frac{1}{2}\ln|a^2+x^2|+c
    [/tex]

    can I do the last step, is this correct ?

    cheers,
    mike
     
  5. Feb 18, 2010 #4

    CompuChip

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    Science Advisor
    Homework Helper

    It is correct, if you differentiate the result you can check that your anti-derivative works. But I don't see how it logically follows, other than "educated guessing."

    If you take my substitution (u = a2 + x2) your integral reduces to the elementary
    [tex]\int \frac{du}{u} = \log|u|[/tex]
    for which no guesswork is required.
     
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