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Integration via Leibniz

  1. Feb 19, 2009 #1
    1. The problem statement, all variables and given/known data
    \int_0^1\frac{x-1}{\ln{x}} dx

    2. Relevant equations
    \Phi(\alpha)=\int_0^1\frac{x^{\alpha}-1}{\ln{x}} dx

    3. The attempt at a solution
    In the answers they say:

    \Phi '(\alpha)=\int_0^1\frac{x^{\alpha}\ln{x}}{\ln{x}} dx=\frac{1}{\alpha+1}

    but the derative is wrong, right? I don't understand how they calculated the derative...
  2. jcsd
  3. Feb 19, 2009 #2


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    The derivative is d/d(alpha). x^alpha=e^(log(x)*alpha). Actually, it is right.
  4. Feb 19, 2009 #3
    You're right I accendentally differentiated w.r.t x, thanks Dick.
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