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RolyPoly

  • 2 years ago

<Integration> \[\frac{1}{5}\int \frac{-x^3+2x^2-3x+4}{x^4-x^3+x^2-x+1}dx\]

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  1. RolyPoly
    • 2 years ago
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    @hartnn If you don't mind giving a hand here...

  2. hobbs
    • 2 years ago
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    @RolyPoly Man i'm sorry but i can't find the answer... ive been tryin for the past 20 mins

  3. SithsAndGiggles
    • 2 years ago
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    I'd suggest partial fractions, but I don't know if that'd get you anywhere with this one.

  4. hobbs
    • 2 years ago
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    @SithsAndGiggles i tried but just got a mess

  5. RolyPoly
    • 2 years ago
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    Where there's a will, there's a way! :D

  6. FoolAroundMath
    • one year ago
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    \(\text{Denominator = } 1-x+x^2-x^3+x^4 = (1+x^5)/(1+x)\) \[ \implies \frac{1}{5}\int\frac{5-(1-x+x^2-x^3+x^4)}{x^5+1}dx=\int\frac{-1}{5(x+1)}+\frac{1}{x^5+1}dx\] Hope this helps

  7. RolyPoly
    • one year ago
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    The problem is I get this integral from \(\int \frac{1}{x^5+1}dx\)... \[\int \frac{1}{x^5+1}dx=\int ( \frac{1}{5(x+1)} + \frac{(-x^3 + 2x^2 - 3x + 4)}{5(x^4-x^3+x^2-x+1)})dx\]

  8. RolyPoly
    • one year ago
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    If you change it back to \(\int \frac{1}{x^5+1}dx\), then, how can I solve this integral?

  9. RolyPoly
    • one year ago
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    @FoolAroundMath

  10. amistre64
    • one year ago
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    not real sure how this would help, but this is an idea im trying to play with \[\frac{\sum_{0}^{4}(-1)^n(4-n)x^n}{\sum_{0}^{4}(-1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(-1)^n4~x^n-\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(-1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(-1)^n4~x^n}{\sum_{0}^{4}(-1)^n~5~x^n}-\frac{\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]

  11. amistre64
    • one year ago
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    when I simply do the division, i get a series representation of:\[\sum_0~(-1)^n(4x^{5n}+x^{5n+1}-x^{5n+2}+x^{5n+3}-x^{5n+4})\]

  12. SithsAndGiggles
    • one year ago
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    @amistre64, I like your idea of using the pattern of coefficients. I also don't know if it helps, but I like the idea nonetheless.

  13. amistre64
    • one year ago
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    thnx, its was either that or try to figure out what the wolf did :) http://www.wolframalpha.com/input/?i=integrate+%284-3x%2B2x%5E2-x%29%2F%285%281-x%2Bx%5E2-x%5E3%2Bx%5E4%29%29+dx I recall hearing that a power series solution is a solution; and if its an important enough solution, they give it a name ;)

  14. satellite73
    • one year ago
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    lol forget it http://www.wolframalpha.com/input/?i= \frac{1}{5}\int+\frac{-x^3%2B2x^2-3x%2B4}{x^4-x^3%2Bx^2-x%2B1}dx

  15. RolyPoly
    • one year ago
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    \[\frac{\sum_{0}^{4}(-1)^n(4-n)x^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]\[=\frac{\sum_{0}^{4}(-1)^n(4)x^n-(-1)^nnx^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]Is this step correct?

  16. satellite73
    • one year ago
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    i am not sure you are going to find a nice closed form for this you can integrate term by term if you expand \(\frac{1}{1+x^5}\) as a power series start with \[\frac{1}{1+x}=1-x+x^2-x^3+...\] and then replace \(x\) by \(x^5\) and get \[\frac{1}{1+x^5}=1-x^5+x^{10}-...\] and i guess you can integrate that term by term

  17. satellite73
    • one year ago
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    gives \[x-\frac{x^6}{6}+\frac{x^{11}}{11}-...\]

  18. RolyPoly
    • one year ago
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    Hmm... Is \[\frac{\sum_{0}^{4}(-1)^n(4-n)x^n}{\sum_{0}^{4}(-1)^n~5~x^n}=\frac{\sum_{0}^{4}(-1)^n(4)x^n-(-1)^nnx^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]correct?

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