Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly

  • 2 years ago

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

  • This Question is Closed
  1. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @hartnn If you don't mind giving a hand here...

  2. hobbs
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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
    Best Response
    You've already chosen the best response.
    Medals 0

    I'd suggest partial fractions, but I don't know if that'd get you anywhere with this one.

  4. hobbs
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @SithsAndGiggles i tried but just got a mess

  5. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where there's a will, there's a way! :D

  6. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\text{Denominator = } 1-x+x^2-x^3+x^4 = (1+x^5)/(1+x)\) \[ \implies \frac{1}{5}\int\frac{5-(1-x+x^2-x^3+x^4)}{x^5+1}dx=\int\frac{-1}{5(x+1)}+\frac{1}{x^5+1}dx\] Hope this helps

  7. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @FoolAroundMath

  10. amistre64
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{\sum_{0}^{4}(-1)^n(4-n)x^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]\[=\frac{\sum_{0}^{4}(-1)^n(4)x^n-(-1)^nnx^n}{\sum_{0}^{4}(-1)^n~5~x^n}\]Is this step correct?

  16. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gives \[x-\frac{x^6}{6}+\frac{x^{11}}{11}-...\]

  18. RolyPoly
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  19. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.