$\int \frac{x^2+1}{x^5-1}dx$
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.
$\int \frac{x^2+1}{x^5-1}dx$
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.
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.
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,
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.
For these cyclotomic denominators, one can get an answer pretty fast by factoring into roots of unity. Let $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, where $\theta_k=2\pi k/5$. Then $\frac{x^2+1}{x^5-1}=\frac{x^2+1}{\prod_{k=0}^4(x-\omega_k)}=\sum_{k=0}^4\frac{A_k}{x-\omega_k}$ Obeserving that $\omega_k^{-k}=\omega_k^{5-k}$ and applying L'Hopital's rule a couple of times, $\begin{align}\lim_{x\rightarrow\omega_k}\frac{(x^2+1)(x-\omega_k)}{x^5-1} & =\lim_{x\rightarrow\omega_k}\frac{(x^2+1)(1)}{5x^4}=\frac15\left(\omega_k^{-2}+\omega_k\right) \\ & =\lim_{x\rightarrow\omega_k}\sum_{j=0}^4\frac{A_j\left(x-\omega_k\right)}{x-\omega_j}=\sum_{j=0}^4A_j\delta_{kj}=A_k\end{align}$ Then $\begin{align}\int\frac{x^2+1}{x^5-1}dx & =\sum_{k=0}^4A_k\int\frac{dx}{x-\omega_k}=\sum_{k=0}^4\frac15\left(\omega_k^{-2}+\omega_k\right)\ln\left(x-\omega_k\right)+C_1 \\ & =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\ln\left(x-\cos\theta_k-i\sin\theta_k\right)+C_1 \\ & =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\left\{\frac12\ln\left(x^2-2x\cos\theta_k+1\right)+i\,\text{atan2}\left(-\sin\theta_k,x-\cos\theta_k\right)\right\}+C_1 \\ & =\sum_{k=0}^4\left(\cos2\theta_k+\cos\theta_k-i\sin2\theta_k+i\sin\theta_k\right)\left\{\frac12\ln\left(x^2-2x\cos\theta_k+1\right)+i\,\text{atan2}\left(x-\cos\theta_k,\sin\theta_k\right)\right\}+C \\ & = \sum_{k=0}^4\left\{\frac12\left(\cos2\theta_k+\cos\theta_k\right)\ln\left(x^2-2x\cos\theta_k+1\right)+\left(\sin2\theta_k-\sin\theta_k\right)\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right\}+C\end{align}$ because the imaginary parts cancel out above. The $k=0$ term is just $\frac25\ln|x-1|$ and the $k=4$ term is a copy of the $k=1$ term, just as the $k=3$ term is a copy of the $k=2$ term. So substituting in the values of the trig functions above and simplifying, we get $\begin{align}\int\frac{x^2+1}{x^5-1}dx=\frac25\ln|x-1| & -\frac1{10}\ln\left(x^2+\frac{1-\sqrt5}2x+1\right) \\ & +\frac{\sqrt{10-2\sqrt5}\left(1-\sqrt5\right)}{20}\tan^{-1}\left(\frac{4x-\sqrt5+1}{\sqrt{10+2\sqrt5}}\right) \\& -\frac1{10}\ln\left(x^2+\frac{1+\sqrt5}2x+1\right) \\ & -\frac{\sqrt{10+2\sqrt5}\left(1+\sqrt5\right)}{20}\tan^{-1}\left(\frac{4x+\sqrt5+1}{\sqrt{10-2\sqrt5}}\right)+C\end{align}$