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Math Help - 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 t=\log_8(x)

    note that \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 \log stands for \ln, most of books use this.
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  4. #4
    Super Member redsoxfan325's Avatar
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    \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx

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

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

    Pull out the ln(8) to get 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 8 gives you -\frac{ln(8)}{log_88} = -\frac{ln(8)}{1}=-ln(8) and plugging in 2 gives you -\frac{ln(8)}{log_82} = -\frac{ln(8)}{\frac{1}{3}} = ln(8)*-3 = -3ln(8).

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

    Thus, \int_{2}^{8} \frac{1}{x(log_8x)^2}\,dx = 6ln(2).
    Last edited by redsoxfan325; March 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 \log stands for \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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