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Cool integral

  1. Apr 29, 2005 #1

    DH

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    Determine the integral:
    [tex]y = \int_{0}^{1} 1/(u^4+1)du [/tex]
    and
    [tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
    and
    [tex]y = \int_{0}^{1} 1/(u^6+1)du [/tex]
     
    Last edited: Apr 29, 2005
  2. jcsd
  3. Apr 29, 2005 #2
    Just factor the denominator (to quadratics) and use partial fractions.
     
  4. Apr 29, 2005 #3

    DH

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    :surprised
    no no, can't do like that!
     
  5. Apr 29, 2005 #4
    don't see why not~
     
  6. Apr 29, 2005 #5

    DH

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    show me the solution!
     
  7. Apr 29, 2005 #6
    [tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

    [tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

    [tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

    and from there it's just minor substitutions to finish.
     
    Last edited: Apr 29, 2005
  8. Apr 29, 2005 #7

    DH

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    Hix, you did the wrong thing form row 1 -> row 2!!!!
     
    Last edited: Apr 29, 2005
  9. Apr 29, 2005 #8

    DH

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    how about this:
    [tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
     
  10. Apr 29, 2005 #9
    Same advice. You can do it yourself this time :wink:
     
  11. Apr 29, 2005 #10

    DH

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    but one more time, can you show me the solution! I really want to know the way you solve it to compare with my own method.... Please give me the solution in detail!
     
  12. Apr 29, 2005 #11
    You can do it almost exactly the same way I did the last one - factor the denominator (in this case, it factors to a product of two irreducible quadractics and the linear factor (x+1)), then use partial fractions to separate it into a sum of functions that you know how to integrate.

    Why don't you post an example of a solution using your method? I'm interested now!
     
    Last edited: Apr 29, 2005
  13. Apr 29, 2005 #12

    DH

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    You made a mistake here:
    [tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

    [tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]
     
    Last edited: Apr 29, 2005
  14. Apr 29, 2005 #13
    indeed I did, the second line should be

    [tex]\int_0^1 \frac{du}{2}\left( \frac{\frac{1}{\sqrt{2}}u + 1}{u^2+\sqrt{2}u+1} - \frac{\frac{1}{\sqrt{2}}u - 1}{u^2-\sqrt{2}u+1}\right),[/tex]

    but the simplification after that is still similar, just with more terms.
     
    Last edited: Apr 29, 2005
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