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Definite Integration - but how to integrate?

  1. May 28, 2008 #1
    Firstly apologies for the calculation, where can I use equation editing software/fonts?

    Show that :
    dx = 1/6ln(7/4) where the limits are : b = 5 and a = 4
    x^2 -9

    I know that if the denominator was x^2 + 9 I could use 1/a tan-1 x/a + C . But do not know how to attempt this - does it involve the substitution of a trig identity, if so which one...? Thank you.
  2. jcsd
  3. May 28, 2008 #2

    Gib Z

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    Partial fractions.

    [tex]\frac{1}{x^2-9} = \frac{A}{x+3} + \frac{B}{x-3}[/tex]. Find A and B, then this integral becomes easy.
  4. May 28, 2008 #3
    Ok right so whenever I see dx as the numerator I presume it's 1? I had resolved this into partial fractions on a previous part of the question, so..
    . 1 .=
    x^2 -9

    . 1 . - . 1 .
    6(x-3) .6(x+3)

    And then with the limits do I substitute values into x of the partial fractions?

    (1/6 ln(2) - 1/6 ln (8)) - (1/6 ln (1) - 1/6ln(7))
    If this is right how do I get to the answer given, or have i missed something completely?
    Many thanks,

    any assistance with the equation writing font/software???
  5. May 28, 2008 #4

    Gib Z

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    With your posts, click "Go Advanced" and just above the text input box, starting with the Bold button, at the end of that row theres a Sigma letter. Click that and play around with it and the preview post =]

    You don't presume its 1, you just realise that [tex]\frac{dx}{f(x)} = \frac{1}{f(x)} dx[/tex]. I'm not sure about your values of x, cross multiply them back and see if it comes back to the original expression. Then make sure you plug your values in correctly, both both integrals. Then you might be able to simplify it with some log identities.
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