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Math Help - Integral question

  1. #1
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    Integral question

    Hi

    I want to integrate the following:

    1/(ln(x)^3) from 2 to infinity.

    Am i allowed to do:

    (ln(x)^-3)
    -3 ln(x)
    pull out -3 and integrate ln (x)?
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  2. #2
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    Quote Originally Posted by Kuma View Post
    Hi

    I want to integrate the following:

    1/(ln(x)^3) from 2 to infinity.

    Am i allowed to do:

    (ln(x)^-3)
    -3 ln(x)
    pull out -3 and integrate ln (x)?


    Of course you're allowed, but be sure the power is on x and NOT on the whole ln(x)...

    Tonio
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Kuma View Post
    1/(ln(x)^3) from 2 to infinity.

    Kuma: even if you don't use \LaTeX code, you can avoid the ambiguous notation writing:

    1 / [ ln (x^3) ] or 1 / [ ( ln x )^3 ]

    which are different functions .
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  4. #4
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    Ah its the entire [ln(x)]^3.
    What do i do in this case? Substitution?
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  5. #5
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    <br />
\displaystyle<br />
\int \frac{dx}{ln^3(x)}=\int \frac{x \ dx}{xln^3(x)}=\int \frac{x \ d(lnx)}{ln^3(x)}=\int x \  \frac{(-1)}{2} \ d \frac{1}{ln^2(x)}=<br />

    <br />
\displaystyle<br />
=-\frac{1}{2} \ \int x \  d \frac{1}{ln^2(x)}<br />

    You may use substitution (I usually don't use it)

    <br />
\displaystyle<br />
u=x ,\ \ v=\frac{1}{ln^2(x)}<br />

    We get

    <br />
\displaystyle<br />
=-\frac{1}{2} \ [ \   x \  \frac{1}{ln^2(x)}-\int \frac{1}{ln^2(x)} \ dx \ ]<br />

    You may continue in the same way.

    Also we may notice x>>1

    <br />
\displaystyle<br />
\frac{1}{ln^3(x)}> \frac{1}{x} \  .<br />
    Last edited by zzzoak; March 28th 2011 at 09:39 AM.
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