Integrating $u^n/(1+u^{2n})$ 
Question: Is there a way to show that$$\int\frac {u^n}{1+u^{2n}}\, du=\frac {u^{n+1}}{n+1}F\left(1,\tfrac {n+1}{2n};\tfrac 32+\tfrac 1{2n};-u^{2n}\right)+C$$

I came upon this when I was generalizing something. Although Wolfram Alpha instantly gives you the answer, it doesn't show you the intermediate steps.
The first idea that came to mind was to try partial fractions, but that will be painstakingly long and tedious. Is there a faster method to get the hypergeometric function and $u^{n+1}$ terms?
 A: The Taylor series for $\frac{u^n}{1+u^{2n}}$ around $u = 0$ is $$\frac{u^n}{1+u^{2n}} = \sum_{k=0}^{\infty} (-1)^ku^{(2k+1)n}$$ which is uniformly convergent on compact subsets of $(-1, 1)$. We can derive this from the Taylor series for $\frac{z}{1+z^2}$ by letting $z = u^n$. Therefore, for $u\in (-1, 1)$, $$\int_0^u \frac{s^n}{1+s^{2n}}\,\mathrm{d}s = \sum_{k=0}^{\infty} \int_0^u (-1)^ks^{(2k+1)n}\,\mathrm{d}s = \sum_{k=0}^{\infty} \frac{(-1)^ku^{1+(2k+1)n}}{1+(2k+1)n}$$ If we let $\operatorname{\Phi}(z, s, a)$ be the Lerch transcendent, then we can rewrite this as $$\int_0^u \frac{s^n}{1+s^{2n}}\,\mathrm{d}s = \frac{u^{n+1}}{2n}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right)$$ If you check page 30 of this book (page 56 of the PDF),$^1$ you can see that this is equivalent to the hypergeometric function given by Wolfram Alpha and Mathematica. To verify that this formula actually holds everywhere the Lerch transcendent is defined, we can use the fact that $$\frac{\partial}{\partial z}\operatorname{\Phi}(z, s, a) = \frac{\operatorname{\Phi}(z, s-1, a)-a\operatorname{\Phi}(z, s, a)}{z}$$ and $\operatorname{\Phi}(z, 0, a) = \frac{1}{1-z}$ to calculate the derivative of $I(u) := \frac{u^{n+1}}{2n}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right)$ with respect to $u$. In particular, we get \begin{align*} I'(u) &= \frac{(n+1)u^n}{2n}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right) \\ &\mathrel{\phantom{=}}+\frac{u^{n+1}}{2n}(-2nu^{2n-1})\left[-\frac{1}{u^{2n}(1+u^{2n})}+\frac{n+1}{2nu^{2n}}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right)\right] \\ &= \frac{(n+1)u^n}{2n}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right)+\frac{u^n}{1+u^{2n}}-\frac{(n+1)u^n}{2n}\operatorname{\Phi}\left(-u^{2n}, 1, \frac{n+1}{2n}\right) \\ &= \frac{u^n}{1+u^{2n}} \end{align*} as expected.

$^1$Erdélyi, A., Magnus, W., Oberhettinger, F., and Tricomi,  F. G., Higher transcendental functions, Vol. I, McGraw-Hill, New York, 1953.
