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I have, for a long period of time, tried to evaluate the indefinite integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$ for different values of $m$, such as $m=2,3,4,6...$, and I have recently thought of generalizing this integral for all values of $m\in\mathbb{N}$. Since evaluating the integral using partial fractions for high values of $m$ by myself is very time consuming, and it happens that I can't find any direct solution on the internet, I tried to look for patterns of the integrals in WolframAlpha. After some comparisons, I found the below expression: $$\int{\dfrac{x^m}{x^{2m}+1}dx}=\sum_{n=1}^{m}(-1)^{n-1}(\dfrac{1}{2m}\sin(\dfrac{(2n-1)\pi}{2m})\ln{|x^2+2\cos(\dfrac{(2n-1)\pi}{2m})x+1|}$$$$+\dfrac{1}{m}\arctan(\csc(\dfrac{(2n-1)\pi}{2m})(x+\cos(\dfrac{(2n-1)\pi}{2m})))\cos(\dfrac{(2n-1)\pi}{2m}))$$ How could I have found the above expression without comparing the integrals?

P.S. It seems that the equivalence $x^{2m}-2x^ma^m\cos(m\theta)+a^{2m}\equiv\displaystyle\prod_{r=0}^{m-1}(x^2-2ax\cos(\theta+\dfrac{2r\pi}{m})+a^2)$ for $a\in\mathbb{R}^+, \theta\in\mathbb{R}, m\in\mathbb{N}$ can be used, yet I don't know how I can evaluate partial fractions with it.

StrongestStrike
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1 Answers1

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Factorize $x^{2n}+1= \prod_{k=1}^{2n}(x-x_k)$ with $x_k= e^{i a_k},\>a_k=\frac{(2k-1)\pi}{2n}$. Then $$\frac{x^n}{x^{2n}+1}=-\frac1{2n}\sum_{k=1}^{2n}\frac{x_k^{n+1}}{x-x_k}$$ and integrate \begin{align} &\int \frac{x^n}{x^{2n}+1}dx \\=& -\frac1{2n}\sum_{k=1}^{2n}\int \frac{x_k^{n+1}}{x-x_k} dx = -\frac1{2n}\sum_{k=1}^{2n} x_k^{n+1} \ln(x-x_k)\\ = &-\frac1{2n} \sum_{k=1}^{2n} [\cos(n+1)a_k +i \sin(n+1)a_k] \ln(x-\cos a_k -i\sin a_k)\\ = &-\frac1{4n} \sum_{k=1}^{2n} \cos(n+1)a_k \ln(x^2-2x\cos a_k+1)+2\sin(n+1)a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k} \end{align}

where only the real part survives.

Quanto
  • 97,352