Indefinite integral of $\int \frac { dx}{x^{2m}+1}, m \in \mathbb R $ I spent a few hours trying to solve this indefinite integral:
$$\int \frac { dx}{x^{2m}+1}, m \in \mathbb R $$
I tried to transform the fraction to partial fractions getting $\int \frac{\imath}{2(x^m+\imath)} {dx} - \int \frac{\imath}{2(x^m-\imath)} {dx}$, but this doesn't help me because we enter in complex analisys, a field that I've no knowledgement and searching for methods that can help me with this make me nuts. I think the solution should involve some sort of induction, so I tried to solve for $ m= 0, 1, 2...$, but this doesn't help me neither. I've no luck looking for some books in the university library that can help me.
Putting the integral in wolframalpha gives me the result:$$ x \space_2F_1(1,\frac{1}{2m};1 + \frac{1}{2m}; -x^{2m}) \color{silver} {+constant}$$where $_2F_1(1,\frac{1}{2m};1 + \frac{1}{2m}; -x^{2m})$ is the hypergeometric function (I'd never hear about this before) .
This reveals to me that this problem is beyond my capabilities. Can some kind soul make a detailed resolution for me? If possible, not going into deep complex analysis.
 A: The Wolfram Alpha expression of the indefinite integral in terms of the Gauss hypergeometric function is more straightforward than it might seem. We formally expand the integrand in powers of $x$ and integrate term-by-term to obtain
$$\int \frac{dx}{1+x^{2m}} = \int\left[1-x^{2m}+x^{4m}-\cdots\right]dx = x-\frac{x^{2m+1}}{2m+1} +\frac{x^{4m+1}}{m+1}-\cdots.$$ This expression is of the form $x \,f(-x^{2m})$ with $$f(u)=1+\frac{u}{2m+1}+\frac{u^2}{4m+1}+\cdots =\sum_{k=0}^\infty \frac{u^k}{2km+1}.$$ To compare with Wolfram Alpha, we recall that the Gauss hypergeometric function is defined as $\displaystyle _2F_1(a,b;c;u)=\sum_{k=0}^\infty \frac{(a)_k(b)_k}{(c)_k }\frac{u^k}{k!}$ where $(a)_k = a(a+1)\cdots (a+k-1)$. The particular case of $(a,b,c)=(1,\frac{1}{2m},1+\frac{1}{2m})$ gives
$$\frac{(1)_k}{k!}\frac{(\frac{1}{2m})_k}{(1+\frac{1}{2m})_k }=\frac{k!}{k!}\frac{\frac{1}{2m}(\frac{1}{2m}+1)\cdots (\frac{1}{2m}+k-1)}{(\frac{1}{2m}+1)(\frac{1}{2m}+2)\cdots (\frac{1}{2m}+k)}=\frac{\frac{1}{2m}}{\frac{1}{2m}+k}=\frac{1}{1+2mk}$$ and we conclude that $f(u)={_2F}_1(1,\frac{1}{2m},1+\frac{1}{2m};u)$ in agreement with Wolfram Alpha.
Note: For the formal expansion in powers of $x$ to be justified, we need $|x^m|<1$. Hence the above reasoning will be valid near $x=0$ under the assumption $m>0$.
A: It only takes complex arithmetic to expand in partial fractions, not complex analysis. There is a huge difference. Let $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, where $\theta_k=\frac{(2k+1)\pi}{2m}$. Then $\omega_k^{2m}=-1$ and
$$\frac1{x^{2m}+1}=\frac1{\prod_{k=0}^{2m-1}}(x-\omega_k)=\sum_{k=0}^{2m-1}\frac{A_k}{x-\omega_k}$$
By L'Hopital's rule,
$$\begin{align}\lim_{x\rightarrow\omega_k}\frac{(x-\omega_k)}{x^{2m}+1}&=\lim_{x\rightarrow\omega_k}\frac1{2mx^{2m-1}}=\frac1{2m\omega_k^{2m-1}}=-\frac{\omega_k}{2m}\\
&=\sum_{\ell=0}^{2m-1}A_{\ell}\lim_{x\rightarrow\omega_k}\frac{x-\omega_k}{x-\omega_{\ell}}=\sum_{\ell=0}^{2m-1}A_{\ell}\delta_{k\ell}=A_k\end{align}$$
And now since $\theta_{2m-1-k}=2\pi-\theta_k$ it follows that
$$\begin{align}\frac1{x^{2m}+1}&=\sum_{k=0}^{2m-1}-\frac{\omega_k}{2m(x-\omega_k)}=-\frac1{2m}\sum_{k=0}^{m-1}\left(\frac{\omega_k}{x-\omega_k}+\frac{\omega_k^*}{x-\omega_k^*}\right)\\
&=-\frac1{2m}\sum_{k=0}^{m-1}\frac{x\omega_k-1+x\omega_k^*-1}{x^2-(\omega_k+\omega_k^*)x+1}=-\frac1{2m}\sum_{k=0}^{m-1}\frac{2x\cos\theta_k-2}{x^2-2x\cos\theta_k+1}\\
&=\frac1{2m}\sum_{k=0}^{m-1}\frac{-2(x-\cos\theta_k)\cos\theta_k+2\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\end{align}$$
So now we can integrate
$$\begin{align}\int\frac{dx}{x^2m+1}&=\frac1{2m}\sum_{k=0}^{m-1}\int\frac{-2(x-\cos\theta_k)\cos\theta_k+2\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}dx\\
&=\frac1{2m}\sum_{k=0}^{m-1}\left[-\cos\theta_k\ln\left|x^2-2x\cos\theta_k+1\right|+2\sin\theta_k\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$
Now, you can simplify the above a little bit because $\cos\theta_{m-1-k}=-\cos\theta_k$ and $\sin\theta_{m-1-k}=\sin\theta_k$ but you have to be a little careful because there is a special term when $\theta_k=\pi/2$ when $m$ is odd and combining the arctangents gets a little tricky because their individual ranges span a width of $\pi$ so combining them in pairs requires a range of $2\pi$ so you would need the $\text{atan2}$ function or a restricted domain for $x$.
