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Misc integral

  1. Mar 30, 2010 #1
    [tex]
    \int \frac{ln(x+1)}{x^2}dx
    [/tex]
    [tex]
    u=x+1
    [/tex]
    [tex]
    \int\frac{lnu}{(u-1)^2}du
    [/tex]
    [tex]
    \int \frac{A}{u-1}+\frac{B}{(u-1)^2}du
    [/tex]
    [tex]
    lnu=A(u-1)+B
    [/tex]
    [tex]
    B=0,A=ln2
    [/tex]
    [tex]
    \int \frac{ln2}{u-1}du
    [/tex]
    [tex]
    ln2ln|x|+C
    [/tex]
     
  2. jcsd
  3. Mar 30, 2010 #2

    rl.bhat

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    Homework Helper

    You have to do integration by parts.
    Int[ln(x+1)/x^2] = ln(x+1)*intg(1/x^2) - Intg{[Intg(1/x^2)*d/dx[ln(x+1)]}
    Now proceed.
     
  4. Mar 30, 2010 #3
    where is the error
     
  5. Mar 30, 2010 #4

    rl.bhat

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    Homework Helper

    I = -1/x*ln(x+1) - int[(-1/x)*1/(x+1)]
    = -1/x*ln(x+1) + int(1/x) - int1/(x+1)
    Now find the integration.
    In your partial factor, there is no ln function in the right hand side. So you can't equate the coefficients.
     
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