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Math Help - Help proving an integration formula via differentiation

  1. #1
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    Help proving an integration formula via differentiation

    I'm stuck on something simple, I'd be grateful for any help.

    My text ("Forgotten Calculus", Bleau) states the following integration formula:

    integral [dx / (x ln x)] = ln |ln x| + C.

    In the course of a problem where I've applied the formula, I realised I can't prove it.

    Here's my attempt, for what it's worth.

    (Re notation:
    exponents are denoted with the "^" sign, e.g x squared = x^2;
    first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))

    f(x) = ln |ln x| + C

    I want to differentiate this via the product rule, so I rewrite the function as

    f(x) = (ln x^0) * (ln x) + C

    Using the product rule,

    f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ]
    = [ (ln 1) * (1/x)] + [ (0/1) * (ln x)]
    = [ ln / x ] + [0]
    = ln / x.

    I don't see how ln / x = ln |ln x|.

    Any advice?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lingyai
    I'm stuck on something simple, I'd be grateful for any help.

    My text ("Forgotten Calculus", Bleau) states the following integration formula:

    integral [dx / (x ln x)] = ln |ln x| + C.

    In the course of a problem where I've applied the formula, I realised I can't prove it.

    Here's my attempt, for what it's worth.

    (Re notation:
    exponents are denoted with the "^" sign, e.g x squared = x^2;
    first derivatives are denoted with the " ' " symbol, e.g. the 1st derivative of f(x) = f'(x))

    f(x) = ln |ln x| + C

    I want to differentiate this via the product rule, so I rewrite the function as

    f(x) = (ln x^0) * (ln x) + C
    \ln |\ln(x)| =\ln(|\ln(x)|),

    you need to use the chain rule.

    RonL
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  3. #3
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    You have, x>0
    \int \frac{dx}{x\ln x}
    Let, u=\ln x then,
    du/dx=1/x
    Thus,
    \int \frac{1}{\ln x} \cdot \frac{1}{x} dx
    Gives after substitution,
    \int \frac{1}{u} \frac{du}{dx} dx=\int \frac{1}{u} du
    Thus,
    \ln|u|+C
    Thus, substitute back,
    \ln|\ln x|+C
    Last edited by ThePerfectHacker; May 22nd 2006 at 03:11 PM.
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