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Omniscience

  • 2 years ago

Integrate (1+x^2)/(1+x^4)dx

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  1. Omniscience
    • 2 years ago
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    \[\large \int\limits\frac{1+x^{2}}{1+x^{4}} dx\]

  2. lgbasallote
    • 2 years ago
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    ugh why couldnt it be 1 - x^4 :p lol

  3. Omniscience
    • 2 years ago
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    If it was it would have defeated the purpose of this integral. (:

  4. lgbasallote
    • 2 years ago
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    exactly why im complaining :p

  5. Omniscience
    • 2 years ago
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    No need to complain. Just think of an intuitive method. :)

  6. blockcolder
    • 2 years ago
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    Completing the square of the denominator then a substitution is what I have in mind, but I'm not completely sure.

  7. Omniscience
    • 2 years ago
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    Not that easy.

  8. Omniscience
    • 2 years ago
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    There are two methods.

  9. goxenul
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=integrate+%5Cfrac%7B1%2Bx%5E%7B2%7D%7D%7B1%2Bx%5E%7B4%7D%7D+dx (look at "Show steps") Well, this looks... difficult. I wonder if there's an easier way to do this, than splitting it into two integrals.

  10. lgbasallote
    • 2 years ago
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    okay...hmm \[u = \frac{1}{1+x^4} \rightarrow du = \frac{4x^3}{(1+x^4)^2}\] \[dv = 1+ x^2 \rightarrow v = \tan^{-!} x\] LOL really isn't easy :P

  11. blockcolder
    • 2 years ago
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    Separation, perhaps? \[\frac{1}{1+x^4}+\frac{x^2}{1+x^4}\] Might this work?

  12. Omniscience
    • 2 years ago
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    Not really. Wolfram Alpha method is tedious.

  13. Omniscience
    • 2 years ago
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    You can try if it works; but I dont think so.

  14. lgbasallote
    • 2 years ago
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    the first term would be integrable i think

  15. Ishaan94
    • 2 years ago
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    \[\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)+2} \]This should do it.

  16. blockcolder
    • 2 years ago
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    Wow. O_O

  17. Ishaan94
    • 2 years ago
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    \[\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)^2+2}\]

  18. Omniscience
    • 2 years ago
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    Very nice Ishaan94; trig sub would work as well (: But that is easier.

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