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

Question

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

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.

Answer 1

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.

Answer 2

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,

Answer 3

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 quite straight forward.