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Math Help - Order of integration

  1. #1
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    Order of integration

    If I evaluate an integral
    \int {1 \over 3x}dx = \ln {3x} + \ln c
    or the constant can be brought outside:  \int {1 \over 3x}dx = {1 \over 3}\int {1 \over x} dx = {1 \over 3} \ln x + \ln c
    which are different answers, how is this?
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  2. #2
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    Quote Originally Posted by bigdoggy View Post
    If I evaluate an integral
    \int {1 \over 3x}dx = \ln {3x} + \ln c
    or the constant can be brought outside:  \int {1 \over 3x}dx = {1 \over 3}\int {1 \over x} dx = {1 \over 3} \ln x + \ln c
    which are different answers, how is this?
    Would it be because:
    \ln 3x + \ln c = \ln 3 + \ln x + \ln c = \ln x + \ln c
    since the 3 is absorbed into the constant?

    The reason I ask is because I've got this : dy = - {du \over 5u} which can be solved to give different answers:
     1) -y = \ln 5u + \ln c \rightarrow u = C \exp^{-y} \  2)\ 5y = - \ln u - \ln c \rightarrow u = C \exp ^{-5y}

    how can this be so?
    Last edited by bigdoggy; August 25th 2009 at 02:26 PM.
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  3. #3
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    Not quite

    int(dx/3x) is not ln(3x) but 1/3ln(3x) +c = 1/3ln(x) + c (1/3ln(3)is absorbed into c

    To see this differentiate ln(3x) -- you don't get 1/3x

    When integrating int(dx/3x) if you leave the 3 in the integral

    you should use substitution u = 3x du = 3dx

    1/3du = dx and you get 1/3int(1/udu)
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