I am a student who is preparing for IIT exam. I was just practicing calculus and encountered this problem. I tried different substitutions but none of them seemed to work. So what is the primitive function of $$\int \frac{1}{x^{2n} +1} \, \mathrm{d}x $$ ?

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    $\begingroup$ There's no simple form which is valid for all $n \in \mathbb{N}$ (you need to use hypergeometric functions). That's why your substitutions lead to nowhere. $\endgroup$
    – Glorfindel
    Commented Jul 8, 2015 at 16:58
  • $\begingroup$ Hmm... it seems the result involves hypergeometric functions... are you sure this is the correct problem? $\endgroup$
    – Cristopher
    Commented Jul 8, 2015 at 16:59
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    $\begingroup$ You can solve that by decomposition in simple fractions, as the roots are easy to find. But the computation is tedious. $\endgroup$
    – user65203
    Commented Jul 8, 2015 at 17:00
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    $\begingroup$ Yah, it is. My teacher told that this is one of the hardest Integration sum for our level. Well I have never heard of hypergeometric functions before. $\endgroup$ Commented Jul 8, 2015 at 17:00
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    $\begingroup$ Yes, you can use the "roots of unity". x^n=k have a set of solutions which are spread on a circle in the complex plane for n even you can probably pair them together to get real coefficients. $\endgroup$ Commented Jul 8, 2015 at 17:06

3 Answers 3


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

$$f(x)=\prod_{k=1}^n(x-x_k)^{-1} \tag {1}$$

where $x_k=e^{i(2k-1)\pi/n}$, $k=1, \cdots,n$.

We can also express $(1)$ as

$$f(x)=\sum_{k=1}^na_k(x-x_k)^{-1} \tag {2}$$

where $a_k=\frac{-x_k}{n}$.

Now, we can write

$$\begin{align} \int\frac{1}{x^n+1}dx=-\frac1n\sum_{k=1}^nx_k\log(x-x_k)+C \end{align}$$

which can be more explicitly written as

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1}{x^n+1}dx=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C'} $$

where $x_{kr}$ and $x_{ki}$ are the real and imaginary parts of $x_k$, respectively, and are given by

$$x_{kr}=\text{Re}\left(x_k\right)=\cos \left(\frac{(2k-1)\pi}{n}\right)$$

$$x_{ki}=\text{Im}\left(x_k\right)=\sin \left(\frac{(2k-1)\pi}{n}\right)$$


The integral of $\frac{1}{1+x^{2n}}$ is a special case for the development herein. Simply let $n\to 2n$.


As requested, we will derive the form $a_k=-\frac{x_k}{n}$. To that end, we use $(2)$ and observe that

$$\begin{align} \lim_{x\to x_\ell}\left((x-x_{\ell})\sum_{k=1}^{n}a_k(x-x_k)^{-1}\right)&=\lim_{x\to x_\ell}\left((x-x_{\ell})\frac{1}{1+x^n}\right) \tag 3 \end{align}$$

The left-hand side of $(3)$ is simply $a_{\ell}$. For the right-hand side, straightforward application of L'Hospital's Rule yields

$$\begin{align} \lim_{x\to x_\ell}\left(\frac{(x-x_{\ell})}{1+x^n}\right)&=\frac{1}{nx_{\ell}^{n-1}} \end{align}$$

Finally, we note that since $x_{\ell}^n=-1$, then

$$\begin{align} \frac{1}{nx_{\ell}^{n-1}}&=\frac{x_{\ell}}{nx_{\ell}^n}\\\\ &=-\frac{x_{\ell}}{n} \end{align}$$

Thus, we have that

$$\bbox[5px,border:2px solid #C0A000]{a_{k}=-\frac{x_k}{n}}$$

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    $\begingroup$ After the factorization given by Dr. MV we see that $|x_k|<1$ so that we may use geometric series such as $$ \int {\frac{{dx}}{{x^{2n} + 1}}} = \int {\sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k x^{2nk} } dx} = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \int {x^{2nk} dx} } = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \frac{{x^{2nk + 1} }}{{2nk + 1}}} $$ $\endgroup$ Commented Jul 8, 2015 at 23:55
  • $\begingroup$ yes, you r right $\endgroup$ Commented Jul 9, 2015 at 0:09
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    $\begingroup$ @mwomath And a finite sum seems more elegant than an infinite series solution. Don't you agree? $\endgroup$
    – Mark Viola
    Commented Jul 9, 2015 at 0:13
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    $\begingroup$ The most interesting part is the computation of the $a_k$'s. How do you get that result ? $\endgroup$
    – user65203
    Commented Jul 9, 2015 at 7:33
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    $\begingroup$ @paraskhosla And that user's assertion that $|x_k|<1$ is not correct. But certainly for $x\in(-1,1)$, that series expansion is valid. $\endgroup$
    – Mark Viola
    Commented Feb 27, 2019 at 18:38

It has no simple closed form, unless you also give some nice integration endpoints, such as $\int \limits _0 ^\infty$. For your curiosity, you get $x \space {}_2 F _1 (\frac 1 {2n}, 1, 1+ \frac 1 {2n}, -x ^{2n})$, where ${}_2 F _1$ is the hypergeometric function.

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    $\begingroup$ By partial fraction decomposition, you get a linear combination of $2n$ terms $\ln(x-z_k)$, where $z_k$ is a $2n^{th}$ root of $-1$, nothing really exotic. $\endgroup$
    – user65203
    Commented Jul 8, 2015 at 17:18
  • $\begingroup$ @YvesDaoust: I feel that the OP hasn't studied complex analysis yet. Plus, even with complex functions, you would still get a sum of simple terms - it depends on your taste whether you call that a closed formula or not. $\endgroup$
    – Alex M.
    Commented Jul 8, 2015 at 17:20
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    $\begingroup$ I do, as long as the sum is finite. $\endgroup$
    – user65203
    Commented Jul 8, 2015 at 18:14

First find $x^{2n}+1 = 0$, then split into sum of fractions $1/(x+b)$ and $1/(x^2+bx+c)$ 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?

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    $\begingroup$ Partial fraction decomposition. All of these roots have nice expression in trigonometric form. So maybe the intended answer is a sum of $n$ things involving trig functions. $\endgroup$
    – GEdgar
    Commented Jul 8, 2015 at 17:05
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    $\begingroup$ Yes, logarithms and arctan and maybe a few others. $\endgroup$ Commented Jul 8, 2015 at 17:13

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