I know how to find the integral below, but I would like to know if there is any clever or general formula for the integral, since my method involves simple polynomial division...
$\int \frac{1}{1+\sqrt[n]x}dx$
Thanks.
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$$\int \frac{1}{1+\sqrt[n]x}dx = \sum_{k=0}^{\infty}(-1)^k \int x^{\frac{k}{n}}dx= \sum_{k=0}^{\infty}(-1)^k \frac {n\,x^{\frac{k}{n}+1}}{k+n} = nx\Phi(-x^\frac{1}{n},1,n), $$
where $\Phi(z,s,\alpha)$ is the LerchPhi function
$$ \Phi(z, s, \alpha) = \sum_{n=0}^\infty \frac { z^n} {(n+\alpha)^s}. $$
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2I've just corrected the sign of 1st argument of $\Phi$ (I hope you don't mind this, it was simpler than explaining where the misprint is). The result can also be written as $x{}_2F_1(1,n,n+1,-x^{1/n})$. – Start wearing purple Jun 04 '13 at 20:00
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@O.L.: Thank you for doing that. I really appreciate it. – Mhenni Benghorbal Jun 04 '13 at 20:05
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@O.L. For hypergeometric solution, I just had noticed that Mathematica wolfram alpha can give this solution, so I wanted to get something different. – Mhenni Benghorbal Jun 04 '13 at 20:10
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@BabakS.: Thanks for the comment. – Mhenni Benghorbal Aug 02 '13 at 14:18