I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C

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  1. Sifr's Avatar
    1 18-04-2012  00:17
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    I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    I need to integrate 1/(xlnx)

    I'm doing so by integration by parts


    1/(xlnx) = 1/x . 1/ lnx

    u = 1/lnx dv/dx = 1/x

    I integrate and use uv - int(du/dx . v) to get 1 which if definitely wrong

    I've looked at how people do it over the net, and they always take u as lnx?
    I cannot for the life of me work out why?

    When we integrate by parts we: int(u.dv/dx)
    why isn't it working for me?

    My u value is 1/lnx and my dv/dx is 1/x Becuase when multiply these two together we get 1/(xlnx)
    Last edited by Sifr; 18-04-2012 at 00:18.
  2. claret_n_blue's Avatar
    2 18-04-2012  00:27
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    You can't integrate ln(x), you can only differentiate it to 1/x so therefore you need to set u = ln(x) so you can differentiate it.
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  3. Sifr's Avatar
    3 18-04-2012  00:29
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by claret_n_blue)
    You can't integrate ln(x), you can only differentiate it to 1/x so therefore you need to set u = ln(x) so you can differentiate it.
    Yes I know that's why I chose u= 1/ lnx ...
    Just don't get why everyone gets u = lnx :/
  4. f1mad's Avatar
    4 18-04-2012  00:31
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Sifr)
    Yes I know that's why I chose u= 1/ lnx ...
    Just don't get why everyone gets u = lnx :/
    Because its differential is 1/x, and you already have 1/x in the integrand (for substitution that is).
    Last edited by f1mad; 18-04-2012 at 00:41.
  5. porkstein's Avatar
    5 18-04-2012  00:32
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    You choose u = lnx so that dx = xdu. This then cancels with the other x on the bottom of the fraction.
  6. Chwirkytheappleboy's Avatar
    6 18-04-2012  00:32
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    Do you have to use integration by parts?

    \dfrac{1}{x\ln{x}}=\dfrac{\frac{ 1}{x}}{\ln{x}}

    What is \displaystyle \int \dfrac{f'(x)}{f(x)}?
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  7. f1mad's Avatar
    7 18-04-2012  00:34
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Chwirkytheappleboy)
    Do you have to use integration by parts?

    \dfrac{1}{x\ln{x}}=\dfrac{\frac{ 1}{x}}{\ln{x}}

    What is \displaystyle \int \dfrac{f'(x)}{f(x)}?
    No you don't; you could use what you have done or substitution too.
  8. claret_n_blue's Avatar
    8 18-04-2012  00:34
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Sifr)
    Yes I know that's why I chose u= 1/ lnx ...
    Just don't get why everyone gets u = lnx :/
    Ohh. Well if you differentiate it you get 1/x and this can be substituted for the 1/x. You don't need to use by parts, you use substitution:

    \int \frac{1}{x \cdot ln(x)} dx

    Let u = ln(x)
    du = (1/x) dx

    \int \frac{1}{x \cdot u} dx = \int \frac{1}{u} du = ln(u) + C = ln (ln x) + C
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  9. Sifr's Avatar
    9 18-04-2012  00:37
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Chwirkytheappleboy)
    Do you have to use integration by parts?

    \dfrac{1}{x\ln{x}}=\dfrac{\frac{ 1}{x}}{\ln{x}}

    What is \displaystyle \int \dfrac{f'(x)}{f(x)}?
    I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me
  10. f1mad's Avatar
    10 18-04-2012  00:43
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Sifr)
    I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me
    Not necessarily, that's why we have various integration methods.

    Consider trying to integrate ln(x).

    This isn't possible via substitution, only by parts.
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  11. porkstein's Avatar
    11 18-04-2012  00:46
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    Well perhaps you could show the working that gave you the answer 1? Because that's not what I get when I do it by parts...
  12. Chwirkytheappleboy's Avatar
    12 18-04-2012  00:58
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Sifr)
    I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me
    So if you're saying u = \dfrac{1}{\ln{x}}, \dfrac{dv}{dx} = \dfrac{1}{x}

    then we have \dfrac{du}{dx} = -\dfrac{1}{x(\ln{x})^2} and v = \ln{x}

    \displaystyle \int \dfrac{1}{x\ln{x}} = 1 + \displaystyle \int \dfrac{1}{x\ln{x}} + c

    Since the second term of the right hand side is the same as the left hand side, we end up with the wonderous result of -c = 1. This is pretty useless, so since it doesn't get us anywhere we say "ok let's try another method". Use the easier way I suggested earlier and then go to bed
  13. SZRoberson's Avatar
    13 18-04-2012  02:09
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    IBP (Integration by Parts) is not necessary in this problem. I encountered this integral in a homework site a long time ago.

    Let's use substitution.

    \displaystyle\int \frac {1}{x\ln x}\ dx is our integral. I see I have a function (the natural logarithm) and its derivative \left \dfrac {1}{x}\right sitting in the integrand. I'll make the following substitution:

    u = ln\ x du = \dfrac {1}{x}\ dx

    Our integrand now becomes \displaystyle\int \frac {1}{u}\ du . This we can do easily, and you should grab ln (ln\ x) + C.
  14. WTSFG's Avatar
    14 18-04-2012  02:15
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    Cos mama said, mama said that if you take it and integrate it, that's what you get
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  15. nuodai's Avatar
    15 18-04-2012  08:42
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Sifr)
    I can use another method, but it's the principle, every method should work? I'm really pedantic and it bugs me
    There's no reason to think that integration by parts should work just because you have a product in your integral. Often the best way to deal with a product is integration by substitution, particularly when one of the terms is a function of a function and the other term is the derivative of the 'inside function'.

    Think about it: integration by parts states

    \displaystyle \int f(x)\cdot \dfrac{dg}{dx}\, dx = f(x)g(x) - \int \dfrac{df}{dx} \cdot g(x)\, dx

    Compare this with the product rule (rearranged):

    f(x) \cdot \dfrac{dg}{dx} = \dfrac{d}{dx}(f(x)g(x)) - \dfrac{df}{dx} \cdot g(x)

    And integration by substitution states

    \displaystyle \int f(u(x))\dfrac{du}{dx}\, dx = \int f(u)\, du

    Compare this with the chain rule:

    \dfrac{df}{du} \dfrac{du}{dx} = \dfrac{d}{dx} f(u(x))

    As you can see, integration by parts reverses the product rule (which you can think of as 'differentiation by parts') and integration by substitution reverses the chain rule (which you can think of as 'differentiation by substitution').

    Both LHS integrals involve products, but what you get out is a very different type of function. Only in special rare cases are both methods appropriate for calculating the antiderivative.
    Last edited by nuodai; 18-04-2012 at 08:44.
  16. TheJ0ker's Avatar
    16 18-04-2012  08:45
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    (Original post by Chwirkytheappleboy)
    Do you have to use integration by parts?

    \dfrac{1}{x\ln{x}}=\dfrac{\frac{ 1}{x}}{\ln{x}}

    What is \displaystyle \int \dfrac{f'(x)}{f(x)}?
    This is the easiest way to do it, writing it like that shows the integral is clearly ln(ln(x)) + c
  17. Philor9294's Avatar
    17 4 Weeks Ago
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    Re: I don't understand WHY the integral of 1/(xlnx) is ln(lnx) + C
    Not sure if you still need an answer, but I got 1/2 x ln(x)2 + c

    used integrating by parts.

    u=lnx u'=1/x
    v'=1/x v=lnx

    integral(1/xlnx)dx = (lnx)2 - integral(1/xlnx)dx
    2integral(1/xlnx)dx = (lnx)2
    integral (1/xlnx)dx = 1/2 x (lnx)2 + c
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