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After An interesting integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$, I want to generalize a similar integral below $$\int{\frac{x^m}{x^{2m}-1}dx}$$ for all values of $m\in\mathbb{N}$. Below are my steps:

$$\int{\frac{x^m}{x^{2m}-1}dx}=\int{\frac{x^m}{(x^m-1)(x^m+1)}dx}=\frac{1}{2}\left(\int{\frac{1}{x^m-1}dx}+\int{\frac{1}{x^m+1}dx}\right)$$

I don't know what my next step should be.

How should I solve this integral?

Quanto
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StrongestStrike
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  • Hi, are you familiar with the Beta function? You can transform the integral $\int\frac{1}{x^n+1}$ into such a form. –  Feb 07 '21 at 16:56
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    You could also write $1/(x^m\pm1)$ as partial fractions using roots of unity. – J.G. Feb 07 '21 at 16:58

1 Answers1

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Fractionalize partially the integrand $$\frac{x^n}{x^{2n}-1}=\sum_{k=1}^{2n}\frac{c_k}{x-x_k}$$ with $x_k= e^{i a_k},\>a_k=\frac{k\pi}{n}$ and apply the L’Hoptital’s rule to obtain the coefficients $$c_k = \lim_{x\to x_k}\frac{x^n(x-x_k)}{x^{2n}-1}=\frac{(-1)^kx_k}{2n}$$

Then \begin{align} &\int \frac{x^n}{x^{2n}-1}dx =\frac1{2n}\sum_{k=1}^{2n}\int \frac{(-1)^kx_k}{x-x_k} dx = \frac1{2n}\sum_{k=1}^{2n} (-1)^kx_k\ln(x-x_k)\\ = &\frac1{2n} \sum_{k=1}^{2n}(-1)^k (\cos a_k +i \sin a_k) \ln(x-\cos a_k -i\sin a_k)\\ = &\frac1{2n} \sum_{k=1}^{2n} (-1)^k\left(\cos a_k \ln\sqrt{x^2-2x\cos a_k+1}+\sin a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k} \right) \end{align}

Quanto
  • 120,125