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Integrating Natural Log

  1. Jul 6, 2007 #1
    I'm having real problems integrating a natural log. The problem I have been set is (where S = integration sign)

    S (x - 2) Ln(3x)

    I'm trying to use the integration by parts rule but keep getting the wrong answer and I think it might be to do with the natural log. I have used

    f(x) = Ln(3x) f'(x) = 1/x g'(x) = (x - 2) and g(x) = 1/2x^2 - 2x

    Using the integration by parts formula, I end up with

    Ln(3x) * x^2/2 - 2x + S1/2x - 2x

    The answer is wrong..... Can someone help me out here.... Where am I going wrong?
  2. jcsd
  3. Jul 6, 2007 #2


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    observe that
    [tex]\int \log(x) dx= -x +x \log(x)[/tex] from integration by parts and
    [tex]\int x\log(x) dx=-x^2/4+ x^2 \log(x)/2[/tex] also from integration by parts

    just check you workings again.
  4. Jul 6, 2007 #3


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    Use parentheses where needed! Like what is 1/2x? Is it (1/2)*x or 1/(2*x)??? But check again what the product f'(x)*g(x) is.
  5. Jul 6, 2007 #4

    D H

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    Watch your parentheses and review the integration by parts rule. Hint: You have a sign error.
  6. Jul 6, 2007 #5
    Am I right in assuming that I am to use the integration by parts method and the values that I have chosen for f(x), f'(x), g'(x) & g(x) are correct?

    Apologies for the parenthesis error.... I'm trying to learn how to use the equation package....
  7. Jul 6, 2007 #6
    Ok, here's a go at using Latex.... Hope this makes what I'm doing a bit clearer...

    Right, From the initial problem

    [tex] \int (x - 2) \ln(3x)[/tex] I have used

    [tex]f(x) = \ln(3x) \ f'(x) = \frac{1}{x} \ g(x) = \frac{1}{2}x^2 - 2x \ g'(x) = (x - 2)[/tex]

    By using the integration by parts formula, I have

    [tex] \ln(3x) \times \frac{1}{2}x^2 -2x - \int \frac{1}{x} \times \frac{1}{2} x^2-2x[/tex]

    Am I right in saying this up to now?
    Last edited: Jul 6, 2007
  8. Jul 6, 2007 #7


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    With the brackets in, it's correct.
    Last edited: Jul 6, 2007
  9. Jul 6, 2007 #8
    So we have

    [tex] \int \frac{1}{x} \left[ \frac{1}{2}x^2 - 2x \right] dx \longrightarrow \frac{1}{4}x^2 - 2x[/tex]

    to give us

    [tex] \ln(3x) \left[ \frac{1}{2}x^2 - 2x \right] - \left[ \frac{1}{4}x^2 - 2x \right] [/tex]

    Does this look ok? I think that maybe this is where I'm going wrong... However, I suspect my brackets were in the wrong place....
    Last edited: Jul 6, 2007
  10. Jul 6, 2007 #9


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    Looks good to me.
  11. Jul 7, 2007 #10
    you know, it seems like you got the help you needed

    but here's a suggestion

    couldn't you distribute (ln 3x)

    then integrate both seperately

    you would still have to use integration by parts

    but it would be easier to deal with
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