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Integration by Parts, 1/x dillema

  1. Feb 1, 2008 #1
    the answer to [tex]\int\frac{1}{x}(dx)[/tex] is lnx +c. but if you do it with integration by parts, you end up with

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

    which comes to 0=1. why does this happen?
  2. jcsd
  3. Feb 1, 2008 #2


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    The derivation of integration by parts comes from the product rule so the (uv) term is actually [tex]\int(uv)'[/tex]. The constant from this integration is usually assumed to be combined with the constant from the other integral, but since the other integrals canceled out we still have a constant term 1 + c = 0.

    Though, this is just a guess, and I could be completely wrong.
  4. Feb 1, 2008 #3
    I don't think it's possible to tell what you did wrong unless you give more details of your calculation. Did you start with something like this

    \frac{1}{x} = \frac{1}{x} (Dx) = D\Big(\frac{1}{x}x\Big) - \Big(D\frac{1}{x}\Big)x = D\Big(\frac{1}{x}x\Big)+\frac{1}{x}?

    But I'm not ending up into any paradox if I perform the integration by parts with this. You just get

    \int_a^b \frac{1}{x}dx = \frac{b}{b} - \frac{a}{a} + \int_a^b\frac{1}{x}dx.
  5. Feb 1, 2008 #4
    im not using a definite integral in my calculation. there aren't any limits "a" and "b"
  6. Feb 1, 2008 #5

    Ben Niehoff

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    Ah, but it's a trick! There is really no such thing as an indefinite integral. The notation

    [tex]\int f(x) \; dx[/tex]

    is really just shorthand for

    [tex]\int_a^x f(t) \; dt[/tex]

    for some arbitrary constant a. If you write it out like this, so that a is consistent, then you will resolve the paradox.

    Alternatively, you could be absolutely pedantic about your arbitrary constant C:

    [tex]\int \left( \frac{d}{dx} f(x) \right) dx = f(x) + C[/tex]

    and you'll note that it is certainly true that

    0 = 1 + C

    for some C.
  7. Feb 1, 2008 #6
    In a sense, the integral functions are definite integrals too. If F(x) is chosen so that its derivative is f(x), then it is

    F(x) = \int_{x_0}^x f(u) du

    with some [itex]x_0[/itex].

    Your problem seems to be, that you are using a formula

    \int fg' dx = fg - \int f'g dx

    without knowing what it means. My advice is that try to understand how the formula is derived, and everything should get clearer.

    It seems somebody was faster than me.
    Last edited: Feb 1, 2008
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