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

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

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