For the first one:
$u = \dfrac{1}{1+Gr^{-\alpha}}, dv = 2r\, dr \Longrightarrow du = \dfrac{Gr^{-\alpha-1}}{\left(1+Gr^{-\alpha}\right)^2 } dr, v = r^2$
$\begin{align*}\displaystyle \dfrac{1}{R^2}\int_0^R \dfrac{2r\, dr}{1+Gr^{-\alpha}} & = \dfrac{1}{R^2}\left( \dfrac{R^2}{1+GR^{-\alpha}} - \int_0^R \dfrac{Gr^{1-\alpha}\, dr}{\left(1+Gr^{-\alpha}\right)^2}\right) \\ & = \dfrac{1}{1+GR^{-\alpha}} - \dfrac{G}{R^2} \int_0^R \dfrac{r^{1-\alpha}\, dr}{\left(1+Gr^{-\alpha}\right)^2}\end{align*}$
Keep using integration by parts to yield an infinite sequence. At this point, hopefully, it will resemble the hypergeometric function you are looking for.