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Derivative and integral of the natural log

  1. Feb 26, 2013 #1
    not really a problem, but more curious

    if we differentiate ln(2x) we get 2/(2x) = 1/x by the chain rule, but if we integrate 1/x we get ln|x|? Could anyone explain why this is the case, thanks.
  2. jcsd
  3. Feb 26, 2013 #2


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    ln(2x) = ln(x) + ln(2)

    Don't forget the constant of integration.

    [itex]\displaystyle \int \frac{1}{x}\,dx=\ln(|x|)+C_0=\ln(|x|)+\ln(2)+C_1\,,\ [/itex] where C0 = C1 + ln(2) .
  4. Feb 26, 2013 #3
    As to the absolute value it is a generalization which works in the case of negative numbers for which the real logarithm [itex] ln(x) [/itex] is undefined. That it applies can be seen by forming the derivative in the separate cases when [itex] x > 0 [/itex] and [itex] x < 0 [/itex] respectively. One should still watch for intervals which include zero for there neither [itex] ln(x) [/itex] nor [itex] ln|x| [/itex] are defined.
  5. Feb 27, 2013 #4
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