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Thread: Integral question

  1. #1
    VkL
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    Integral question

    How do you do this?


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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by VkL View Post
    How do you do this?


    Just substitute $\displaystyle t=\log_8(x)$

    note that $\displaystyle \log_8(x)=\frac{\ln(x)}{\ln(8)}$ (may be useful for the dt)
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  3. #3
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    I'm pretty sure that $\displaystyle \log$ stands for $\displaystyle \ln,$ most of books use this.
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  4. #4
    Super Member redsoxfan325's Avatar
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    $\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx$

    If you let $\displaystyle t=log_8x$, then $\displaystyle dt=\frac{1}{xln(8)}$.

    Thus you have $\displaystyle \int_{2}^{8}\frac{ln(8)\,dt}{t^2}$.

    Pull out the $\displaystyle ln(8)$ to get $\displaystyle ln(8)\int_{2}^{8}\frac{\,dt}{t^2} = ln(8)*-\frac{1}{t} = ln(8)*-\frac{1}{log_8x} = -\frac{ln(8)}{log_8x}$.

    Plugging in $\displaystyle 8$ gives you $\displaystyle -\frac{ln(8)}{log_88} = -\frac{ln(8)}{1}=-ln(8)$ and plugging in $\displaystyle 2$ gives you $\displaystyle -\frac{ln(8)}{log_82} = -\frac{ln(8)}{\frac{1}{3}} = ln(8)*-3 = -3ln(8)$.

    Subtracting gives: $\displaystyle -ln(8)+3ln(8) = 2ln(8) = 2ln(2^3) = 6ln(2)$.

    Thus, $\displaystyle \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx = 6ln(2)$.
    Last edited by redsoxfan325; Mar 2nd 2009 at 12:19 PM.
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  5. #5
    Moo
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    Quote Originally Posted by Krizalid View Post
    I'm pretty sure that $\displaystyle \log$ stands for $\displaystyle \ln,$ most of books use this.
    With a base ?
    I don't think so ^^

    Also, it's not most of books which use it ><
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