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How to solve this integral

  1. Jun 16, 2009 #1
    Can you help me solve this one :

    [tex]\int[/tex] [tex]\frac{x+8}{x-8}[/tex]dx
     
  2. jcsd
  3. Jun 16, 2009 #2

    Hootenanny

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    What have you tried thus far?
     
  4. Jun 16, 2009 #3
    Do you know how to do [itex]\int{\log{x} \, dx}[/itex]?
     
  5. Jun 16, 2009 #4
    (x+8)/(x-8) = (x + 8 - 16 + 16)/(x-8) = (x-8 + 16)/(x-8) = 1 + 16/(x-8).

    That's not too hard to integrate, right?
     
  6. Jun 17, 2009 #5
    Sorry guys i have submited the integral without puting the exponent for each X(thats why this one is too simple).

    The one i am trying to solve is :

    [tex]\int[/tex] [tex]\frac{x^3+8}{x^3-8}[/tex]dx
     
  7. Jun 17, 2009 #6
    I have tryed irracional mode (thats the one it requires but i am stuck at the end).
     
  8. Jun 17, 2009 #7

    Hootenanny

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    Why don't you show us what you've done and where your stuck then we can help you?
     
  9. Jun 17, 2009 #8
    Hi Penultimate!

    Alright, let's help you out here. Firstly, can you break up the denominator into a linear factor multiplied by a quadratic expression?
     
  10. Jun 17, 2009 #9
    Well (x^3 + 8)/(x^3 - 8) = (x^3 - 8 + 16)/(x^3 - 8) = 1 + 16/(x^3 - 8).

    x^3-8 can easily be factored, and so use partial fractions on that integral.
     
  11. Jun 17, 2009 #10

    HallsofIvy

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    Actually, [itex]x^3+ 8[/itex] can be factored as easily as [itex]x^3- 8[/itex]. In fact, [itexs]x^n+ a^n[/itex] can be factored easily for all odd n.
     
  12. Jun 17, 2009 #11
    [tex]x^3+8=x^3+2^3=(x+2)(x^2-2x+4)[/tex]

    [tex]x^3-8=x^3-2^3=(x-2)(x^2+2x+4)=(x-2)(x+2)^2[/tex]

    Another way is:
    [tex]\frac{x^3+8}{x^3-8}=\frac{x^3-8+16}{x^3-8}=1+\frac{16}{x^3-8}[/tex]

    Regards.
     
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