Closed form of this integral

Question

After putting values into Wolfram for the integral $$\int_{2}^\infty \frac{x^k}{1 - x^{2k}} dx$$ And it seems that it can return the values. If we have this, how might one go about solving $$\int_{2}^\infty \frac{x^{k-1}}{1 - x^{2k}} dx$$

Answer 1

$$ \int_{2}^{+\infty}\frac{x^k}{1-x^{2k}}\,dx = -\int_{2}^{+\infty}\frac{x^k}{x^{2k}-1}\,dx = -\int_{0}^{1/2}\frac{y^{k-2}}{1-y^{2k}}\,dy $$ can be computed in explicit terms for any integer $k\geq 2$ by partial fraction decomposition, and it equals $$ -\int_{0}^{1/2}\sum_{n\geq 0} y^{(2n+1)k-2}\,dy =-\sum_{n\geq 0}\frac{1}{\left[(2n+1)k-1\right]2^{(2n+1)k-1}}$$ (which is a hypergeometric ${}_2 F_1$ function evaluated at $4^{-k}$) for any $k>1$.

The situation for $\int_{2}^{+\infty}\frac{x^{k-1}}{1-x^{2k}}\,dx$ is much simpler: this is just $-\frac{1}{k}\text{arctanh}(2^{-k})$, for any $k>0$.