Kelly Nelson

2022-01-03

Solve the integral:

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

Elaine Verrett

Beginner2022-01-04Added 41 answers

We have $f\left(x\right)=\frac{1}{{x}^{n}+1}$ . Note that we can write

$f\left(x\right)=\prod _{k=1}^{n}{(x-{x}_{k})}^{-1}$ (1)

where${x}_{k}={e}^{i(2k-1)\frac{\pi}{n}}\text{}k=1,\dots ,n$

We can also express (1) as

$f\left(x\right)=\sum _{k=1}^{n}{a}_{k}{(x-{x}_{k})}^{-1}$ (2)

where$a}_{k}=\frac{-{x}_{k}}{n$

Now, we can write

$\int \frac{1}{{x}^{n}+1}dx=-\frac{1}{n}\sum _{k=1}^{n}{x}_{k}\mathrm{log}(x-{x}_{k})+C$

which can be more explicitly written as

$\int \frac{1}{{x}^{n}+1}dx=-\frac{1}{n}$

$\sum _{k=1}^{n}(\frac{1}{2}{x}_{kr}\mathrm{log}({x}^{2}-2{x}_{kr}x+1)-{x}_{ki}\mathrm{arctan}\left(\frac{x-{x}_{kr}}{{x}_{ki}}\right))+{C}^{\prime}\mid$

where$x}_{kr$ and $x}_{ki$ are the real and imaginary parts of $x}_{k$ respectively, and are given by

${x}_{r}=Re\left({x}_{k}\right)=\mathrm{cos}\left(\frac{(2k-1)\pi}{n}\right)$

${x}_{ki}=Im\left({x}_{k}\right)=\mathrm{sin}\left(\frac{(2k-1)\pi}{n}\right)$

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

$\underset{x\to {x}_{l}}{lim}(div>$

where

We can also express (1) as

where

Now, we can write

which can be more explicitly written as

where

Note 1:

The integral of

Note 2:

As requested, we will derive the form

0

Tiefdruckot

Beginner2022-01-05Added 46 answers

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 ${x}_{2}{F}_{1}(\frac{1}{2n},1,1+\frac{1}{2n},-{x}^{2n})$ , where $2{F}_{1}$ is the hypergeometric function.

karton

Expert2022-01-11Added 613 answers

First find

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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