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anonymous
 3 years ago
<Integration>
\[\frac{1}{5}\int \frac{x^3+2x^23x+4}{x^4x^3+x^2x+1}dx\]
anonymous
 3 years ago
<Integration> \[\frac{1}{5}\int \frac{x^3+2x^23x+4}{x^4x^3+x^2x+1}dx\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@hartnn If you don't mind giving a hand here...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RolyPoly Man i'm sorry but i can't find the answer... ive been tryin for the past 20 mins

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'd suggest partial fractions, but I don't know if that'd get you anywhere with this one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles i tried but just got a mess

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Where there's a will, there's a way! :D

FoolAroundMath
 3 years ago
Best ResponseYou've already chosen the best response.0\(\text{Denominator = } 1x+x^2x^3+x^4 = (1+x^5)/(1+x)\) \[ \implies \frac{1}{5}\int\frac{5(1x+x^2x^3+x^4)}{x^5+1}dx=\int\frac{1}{5(x+1)}+\frac{1}{x^5+1}dx\] Hope this helps

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The problem is I get this integral from \(\int \frac{1}{x^5+1}dx\)... \[\int \frac{1}{x^5+1}dx=\int ( \frac{1}{5(x+1)} + \frac{(x^3 + 2x^2  3x + 4)}{5(x^4x^3+x^2x+1)})dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you change it back to \(\int \frac{1}{x^5+1}dx\), then, how can I solve this integral?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3not real sure how this would help, but this is an idea im trying to play with \[\frac{\sum_{0}^{4}(1)^n(4n)x^n}{\sum_{0}^{4}(1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(1)^n4~x^n\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(1)^n4~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\frac{\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3when I simply do the division, i get a series representation of:\[\sum_0~(1)^n(4x^{5n}+x^{5n+1}x^{5n+2}+x^{5n+3}x^{5n+4})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.3thnx, its was either that or try to figure out what the wolf did :) http://www.wolframalpha.com/input/?i=integrate+%2843x%2B2x%5E2x%29%2F%285%281x%2Bx%5E2x%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 ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol forget it http://www.wolframalpha.com/input/?i= \frac{1}{5}\int+\frac{x^3%2B2x^23x%2B4}{x^4x^3%2Bx^2x%2B1}dx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sum_{0}^{4}(1)^n(4n)x^n}{\sum_{0}^{4}(1)^n~5~x^n}\]\[=\frac{\sum_{0}^{4}(1)^n(4)x^n(1)^nnx^n}{\sum_{0}^{4}(1)^n~5~x^n}\]Is this step correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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 \[\frac{1}{1+x}=1x+x^2x^3+...\] and then replace \(x\) by \(x^5\) and get \[\frac{1}{1+x^5}=1x^5+x^{10}...\] and i guess you can integrate that term by term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gives \[x\frac{x^6}{6}+\frac{x^{11}}{11}...\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... Is \[\frac{\sum_{0}^{4}(1)^n(4n)x^n}{\sum_{0}^{4}(1)^n~5~x^n}=\frac{\sum_{0}^{4}(1)^n(4)x^n(1)^nnx^n}{\sum_{0}^{4}(1)^n~5~x^n}\]correct?
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