If $f(x)$ is polynomial then $\bar{f}(x)$ is obtained from $f$ by changing the sign of coefficient $x$.
I was trying to find a pattern in the anti derivative of $\frac{1}{1+x^m}$ and I noticed an interesting pattern when $m$ is a multiple of $4$ or equivalently $m=4n, \ n \in \mathbb{N}$.
--
First Observation: For $n \in \mathbb{N}$ it seems that there always exist second degree polynomials $p_{1,n}, p_{2,n}, p_{3,n}, \dots, p_{n,n}$ $a_{k,n}$ the the coefficient of $x$ in $p_{k,n}$ and first degree polynomials $q_{1,n}, q_{2,n}, q_{3,n}, \dots, q_{n,n}$ be with $b_{k,n}$ the the coefficient of $x$ in $q_{k,n}$ then $$\int \frac{dx}{1+x^{4n}}= \sum_{k=1}^n \left(a_{k,n}\frac{\ln(p_{k,n}(x))-\ln(\bar{p}_{k,n}(x))}{8n}\right)+\sum_{k=1}^n \left(\frac{\arctan(q_{k,n}(x))-\arctan(\bar{q}_{k,n}(x))}{4b_{k,n}n}\right)$$
Second observation:
It seems to be that $b_{n,k}=\frac{2}{a_{k,n}}$ I also believe that $q_{k,n}$ can be obtained from $p_{k,n}$ but I couldn't find a way to obtain $q_{n,k}$'s absolute term.
Questions
- Is there a way to obtain $q_{n,k}$ from $p_{n,k}$?
- How to prove that for all $n \in \mathbb{N}$ there exist second degree polynomial $p_{1,n}, p_{2,n}, p_{3,n}, \dots, p_{n,n}$ $a_{k,n}$ the the coefficient of $x$ in $p_{k,n}$ and there exist a first degree polynomial $q_{1,n}, q_{2,n}, q_{3,n}, \dots, q_{n,n}$ be with $b_{k,n}$ the the coefficient of $x$ in $q_{k,n}$ then $$\int \frac{dx}{1+x^{4n}}= \sum_{k=1}^n \left(a_{k,n}\frac{\ln(p_{k,n}(x))-\ln(\bar{p}_{k,n}(x))}{8n}\right)+\sum_{k=1}^n \left(\frac{\arctan(q_{k,n}(x))-\arctan(\bar{q}_{k,n}(x))}{4b_{k,n}n}\right)$$
- How to find $p_{n,k}$ and $q_{n,k}$?
