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$$\int \frac{x^2+1}{x^5-1}dx$$

I am unable to integrate it, nothing works. Yes, I can use partial fraction but who remembers factorization of $x^5-1$, I need a better way of doing this.

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You don't have to "remember" the factorization of $x^5-1$. Notice that $x=1$ is a solution and use long division to see that $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. –  chris Apr 14 '12 at 6:55
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@chris: Long division? Geometric sum formula bro. –  anon Apr 14 '12 at 6:56
    
@anon: That works too. It's useful to know that, in general, if you can spot a solution then long division is a way to get the factorization of the polynomial though. –  chris Apr 14 '12 at 6:57
    
If this is homework problem, you have a cruel instructor. (Unless there is a smart way of doing this that I haven't figured out yet) –  Shitikanth Apr 14 '12 at 7:13
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Very cruel indeed; this results in an unholy mass of arctangents and logarithms... –  J. M. Apr 14 '12 at 7:47

3 Answers 3

Take the denominator, find the 5 complex roots,one of them is 1.Descompose the denominator as: $$(x^5 - 1) = (x-1)\cdot(x-a)(x-a^*)\cdot(x-b)(x-b^*)$$ Remember that $(x-a)(x-a^*)$ is a polynomial of second degree with real coefficients. Then apply descomposition in simple (partial) fractions.

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+1: In case the OP or somebody else has problems finding the roots, they are the fifth primitive roots of unity: $a=e^{2\pi i/5}$, $b=e^{4\pi i/5}$. Here $$ (x-a)(x-a^*)=x^2-(a+a^*)x+aa^*=x^2-(2\cos\frac{2\pi}5)x+1. $$ Remember that $$2\cos\frac{2\pi}5=\frac{\sqrt{5}-1}2.$$ The coefficients become a bit cumbersome, but this allows the use of the common techniques of finding the indefinite integral of a rational function. –  Jyrki Lahtonen Apr 14 '12 at 7:53

There's always this solution that can be used in a neighborhood of $0$ with a convergence radius of $1$ : $$ \int \frac{x^2 + 1}{x^5-1} \, dx = \int -(x^2+1) \left( \sum_{i=0}^{\infty} x^{5i} \right) \, dx = \sum_{i=0}^{\infty} \frac{x^{5i+1}}{5i+1} + \sum_{i=0}^{\infty} \frac{x^{5i+3}}{5i+3} $$ but unless that's what you were expecting, I don't thing it's worth very much. Note that the expansion alpha.Debi was suggesting will probably involve at some point the integration of terms of the form $1/(x-a)$, which will most probably bring up logarithms, which are not well-behaving (in the sense that they are multivalued functions over $\mathbb C$) for integration with respect to a path (even though it does work, I'm just mentioning "there's a point to notice there"). Perhaps the solution in terms of logarithms obtained in this fashion is equivalent to this one in a neighborhood of $0$.

Hope that helps,

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But in the descomposition I suggested , we forget the complex numbers after multiplying (x-a)(x-conjugate a),then we continue only in the way of integral of simple fractions all with real coefficients. –  alpha.Debi Apr 14 '12 at 7:51

I'd like to point out that you can factorize $x^5 -1$ quite easily by observing that you are actually finding the fifth roots of 1: $$x =\sqrt[5]{1} = e^{2\pi i k/5}$$ So that as @alpha.Debi pointed out, factorizing the denominator may be long, but is quit straight forward.

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