Cool integral

  1. DH

    DH 22

    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. Just factor the denominator (to quadratics) and use partial fractions.
     
  4. DH

    DH 22

    :surprised
    no no, can't do like that!
     
  5. don't see why not~
     
  6. DH

    DH 22

    show me the solution!
     
  7. [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. DH

    DH 22

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

    DH 22

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

    DH 22

    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. 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. DH

    DH 22

    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. 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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