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Integrate 1/x = ln(x) vs ln(kx)

  1. Feb 21, 2015 #1
    If [itex]k[/itex] is a constant, I know
    [itex]\frac{d}{dx} \ln(x) = \frac{1}{x}[/itex]
    [itex]\frac{d}{dx} \ln(kx) = \frac{k}{kx} = \frac{1}{x}[/itex]

    However, what about [itex]\int\frac{1}{x}[/itex].
    I've been taught to use [itex]\int\frac{1}{x} = \ln(x)[/itex],
    but wouldn't [itex]\int\frac{1}{x} = \ln(kx)[/itex] work as well.
    And if this is true, there are an infinite number of integrals!?!?
     
  2. jcsd
  3. Feb 21, 2015 #2
    Oh god, I just figured it out, right after posting this. I swear I was confounded for hours before this.
    [itex]ln(kx)[/itex] works as well.
    [itex]ln(kx) = ln(x) + ln(k)[/itex] and the [itex]ln(k)[/itex] would cancel out in any definite integrals.
    But that would still mean there are an infinite number of indefinite integrals, which is a bit strange but perhaps possible.
     
  4. Feb 21, 2015 #3

    Mentallic

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    Well, yes!

    ln(kx) = ln(k)+ln(x)

    and since k is just some constant, then ln(k) is also a constant. Remember that when you take the integral, you have to add +c at the end to denote that you can add any constant to the result and still have a correct solution.

    So

    [tex]\int\frac{1}{x}dx=\ln(kx)+c = \ln(x)+\ln(k)+c[/tex]

    for any constant c, but since c is any constant, then we can merge [itex]\ln(k)+c[/itex] into a new constant, say, c2 which is the exact same result as if we left it as ln(x)+c to begin with.
     
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