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Integration-Explanation Please?

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem asks me to integrate this: [itex] ∫(1/x) * (e^{-2 log [3, x]})) dx [/itex]
    I don't know how to write log base 3 x, so that is how I entered it. Sorry if that was confusing. :S

    2. Relevant equations

    3. The attempt at a solution

    The correct answer is [itex] (-log(3))/(2x^{2/(log 3)}) + C [/itex]

    But I have no clue how that was even the answer. Guidance is much appreciated. Thank you.
  2. jcsd
  3. May 24, 2013 #2


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    Rewrite ##\log_3 x## in terms of the natural logarithm: $$\log_ax = \frac{\ln x}{\ln a}$$
  4. May 24, 2013 #3
    So I'd get int (1/x) * (e^(-2 lnx/ln3)) dx? And is the integral of e to any power that expression times the exponent?
  5. May 25, 2013 #4


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    Only if the exponent is a constant (i.e., does not contain any x terms).

    You'll simplify things by using the fact that

    eloge(x) = x
  6. May 25, 2013 #5


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    No. You have this $$\int \frac{1}{x} \exp \left(-2 \frac{\ln x}{\ln 3}\right)\,dx$$ There is an obvious substitution you can make now.
  7. May 25, 2013 #6
    I'm not sure if you can take the integral of 1/x and multiply it by the integral of the ln part?
  8. May 25, 2013 #7


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    No, you can't.

    You have to use some additional properties of logarithms and exponential functions to get a somewhat simple integrand.

    I say somewhat simple because it will just be x to a power -- but the power is pretty weird looking.
  9. May 25, 2013 #8
    Oh! I see. Thank you.
  10. May 25, 2013 #9


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    Since 1/x is the derivative of ln x, you might try the substitution u = ln x as well.
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