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

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RolyPolyBest ResponseYou've already chosen the best response.0
@hartnn If you don't mind giving a hand here...
 9 months ago

hobbsBest 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
 9 months ago

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

hobbsBest ResponseYou've already chosen the best response.0
@SithsAndGiggles i tried but just got a mess
 9 months ago

RolyPolyBest ResponseYou've already chosen the best response.0
Where there's a will, there's a way! :D
 9 months ago

FoolAroundMathBest 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
 9 months ago

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

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

amistre64Best ResponseYou've already chosen the best response.3
not real sure how this would help, but this is an idea im trying to play with \[\frac{\sum_{0}^{4}(1)^n(4n)x^n}{\sum_{0}^{4}(1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(1)^n4~x^n\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\] \[\frac{\sum_{0}^{4}(1)^n4~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\frac{\sum_{0}^{4}n~x^n}{\sum_{0}^{4}(1)^n~5~x^n}\]
 9 months ago

amistre64Best ResponseYou've already chosen the best response.3
when I simply do the division, i get a series representation of:\[\sum_0~(1)^n(4x^{5n}+x^{5n+1}x^{5n+2}+x^{5n+3}x^{5n+4})\]
 9 months ago

SithsAndGigglesBest 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.
 9 months ago

amistre64Best ResponseYou've already chosen the best response.3
thnx, 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 ;)
 9 months ago

satellite73Best ResponseYou've already chosen the best response.0
lol 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
 8 months ago

RolyPolyBest 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?
 8 months ago

satellite73Best ResponseYou've already chosen the best response.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 \[\frac{1}{1+x}=1x+x^2x^3+...\] 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
 8 months ago

satellite73Best ResponseYou've already chosen the best response.0
gives \[x\frac{x^6}{6}+\frac{x^{11}}{11}...\]
 8 months ago

RolyPolyBest ResponseYou've already chosen the best response.0
Hmm... 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?
 8 months ago
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