Kelly Nelson

2022-01-03

Solve the integral:
$\int \frac{1}{{x}^{2n}+1}dx$

Elaine Verrett

We have $f\left(x\right)=\frac{1}{{x}^{n}+1}$. Note that we can write
$f\left(x\right)=\prod _{k=1}^{n}{\left(x-{x}_{k}\right)}^{-1}$ (1)
where
We can also express (1) as
$f\left(x\right)=\sum _{k=1}^{n}{a}_{k}{\left(x-{x}_{k}\right)}^{-1}$ (2)
where ${a}_{k}=\frac{-{x}_{k}}{n}$
Now, we can write
$\int \frac{1}{{x}^{n}+1}dx=-\frac{1}{n}\sum _{k=1}^{n}{x}_{k}\mathrm{log}\left(x-{x}_{k}\right)+C$
which can be more explicitly written as
$\int \frac{1}{{x}^{n}+1}dx=-\frac{1}{n}$
$\sum _{k=1}^{n}\left(\frac{1}{2}{x}_{kr}\mathrm{log}\left({x}^{2}-2{x}_{kr}x+1\right)-{x}_{ki}\mathrm{arctan}\left(\frac{x-{x}_{kr}}{{x}_{ki}}\right)\right)+{C}^{\prime }\mid$
where ${x}_{kr}$ and ${x}_{ki}$ are the real and imaginary parts of ${x}_{k}$ respectively, and are given by
${x}_{r}=Re\left({x}_{k}\right)=\mathrm{cos}\left(\frac{\left(2k-1\right)\pi }{n}\right)$
${x}_{ki}=Im\left({x}_{k}\right)=\mathrm{sin}\left(\frac{\left(2k-1\right)\pi }{n}\right)$
Note 1:
The integral of $\int \frac{1}{1+{x}^{2n}}$ is a special case for the development herein. Simply let $n\to 2n$
Note 2:
As requested, we will derive the form ${a}_{k}=-\frac{{x}_{k}}{n}$. To that end, we use (2) and observe that

Tiefdruckot

It has no simple closed form, unless you also give some nice integration endpoints, such as ${\int }_{0}^{\mathrm{\infty }}$. For your curiosity, you get ${x}_{2}{F}_{1}\left(\frac{1}{2n},1,1+\frac{1}{2n},-{x}^{2n}\right)$, where $2{F}_{1}$ is the hypergeometric function.

karton

First find ${x}^{2n}+1=0$, then split into sum of fractions $1/\left(x+b\right)$ and $1/\left(x+b\right)$ and $1/\left({x}^{2}+bx+c\right)$ and integrate those. I seem to have forgotten what it's called in English. Not partial integration or integration by parts. Partial fraction decomposition maybe?