## anonymous 3 years ago <Integration> $\frac{1}{5}\int \frac{-x^3+2x^2-3x+4}{x^4-x^3+x^2-x+1}dx$

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

@SithsAndGiggles i tried but just got a mess

5. anonymous

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

6. FoolAroundMath

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

7. anonymous

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. anonymous

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

9. anonymous

@FoolAroundMath

10. amistre64

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

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. anonymous

@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

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. anonymous

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. anonymous

$\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. anonymous

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. anonymous

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

18. anonymous

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?