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Need Help with Integral question

  1. Apr 10, 2005 #1
    How do you evaluate this integral:

    integral(1/(x^2-6x+8)

    i dont kno how to subsitute on this or anything. im completely stuck
     
  2. jcsd
  3. Apr 10, 2005 #2

    dextercioby

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    It's standard "arctan" type.To see it,try to write the denominator as a sum of squares...

    Daniel.
     
  4. Apr 10, 2005 #3
    Standard Arctan? I did this on Mathematica's online integral calculator and got the difference of two logarithms. I would do this integral by partial fractions.

    [tex]\int\frac{1}{x^2-6x+8}dx = \int\frac{A}{x-4} + \frac{B}{x-2}dx[/tex]
     
  5. Apr 10, 2005 #4

    dextercioby

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    Sorry.:redface:


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

    and then the substituion [itex] x-3=u [/itex]

    which would give

    [tex] \int \frac{du}{u^{2}-1} [/tex]

    which is typically "arctanh"...

    Daniel.
     
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