computing an integral (fraction of real powers) I want compute the following integral depending on the real parameters $\alpha, \beta > 0$ and $C >0$
$$ \int_0^1 \frac{u^{2\beta}}{C+u^{2(\alpha+\beta)}} du$$
Thanks a lot for any clue !

Consider 
\begin{align}
I(\alpha, \beta) &= \int_{0}^{1} \frac{u^{2\beta}}{c+u^{2(\alpha + \beta)}} \, du \\
&= \frac{1}{c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \int_{0}^{1} u^{2\beta + 2 n (\alpha + \beta)} \, du \\
&= \frac{1}{2c} \, \sum_{n=0}^{\infty} \frac{1}{c^{n}} \, \frac{1}{ n \alpha + (n+1) \beta} \\
&= \frac{1}{2c} \, \sum_{n=0}^{\infty} \frac{1}{\beta + (\alpha + \beta) \, n} \, \frac{1}{c^{n}}
\end{align}
This expression is better used for $c$ being large. 
For $c$ being small the integral can be done in the following way.
\begin{align}
I &= \int_{0}^{1} u^{1 - 2\alpha} \, \partial_{u}\left( \frac{\ln(c+u^{2(\alpha + \beta)})}{2(\alpha + \beta)} \right) \, du \\
&= \ln c + \frac{2 \alpha -1}{2(\alpha + \beta)} \, \int_{0}^{1} u^{-2 \alpha} \, \ln(c + u^{2(\alpha + \beta)}) \, du \\
&= \ln c + \frac{2 \alpha -1}{2(\alpha + \beta)} \, \int_{0}^{1} u^{-2 \alpha} \, \left[ \ln(c)  +  \ln\left(1 + \frac{1}{c} \, u^{2(\alpha + \beta)} \right) \right] \, du \\
\end{align}
