This is an article that I've been studing for a while. I came across this integral equation $$\mathrm dm=\frac{\mathrm dr}{{r}^{z+1}\left[1-{\left(\dfrac{r_h}{r}\right)}^{n+z-1}\right]}$$ where $n$, $z$ and $r_h$ are some parameters and $r$ and $m$ are my variables. The general solution for $n+z>1$ is $$\ m=\frac{-{r}^{-z}}{z} _2F_1\big[1,\frac{z}{n+z-1};1+\frac{z}{n+z-1};{(\frac{r_h}{r})}^{n+z-1}\big]. $$
I want to know how that integral leads to a hypergeometric function of the second kind.
The integral can transformed to an expression that's easier to relate to ${_2}F_1$. First, by a change of variables $r\to x\equiv r_h/r$ and, second, by another change $x\to y\equiv x^z$, the primitive would be given by $$\int \frac{\mathrm d r}{r^{z+1}\left[1-\left(\frac{r_h}{r}\right)^{n+z-1}\right]}=-\frac{1}{r_h^z}\int\frac{x^{z-1}\mathrm d x}{1-x^{n+z-1}}=-\frac{1}{z\,r_h^z}\int\frac{\mathrm dy}{1-y^\alpha},$$ where I took the freedom to define $\alpha=(z+n-1)/z$. The last integral is $$\int\frac{\mathrm d y}{1-y^\alpha}=y\cdot {{_2}F_1}\left(1,\frac1\alpha;1+\frac1\alpha;y^\alpha\right)$$ which, upon undoing of the changes of variables, gives you the result you show for $m$.
Now, if you're interested in how the last integral is related to ${_2}F_1$ you should probably dig into the definition of the hypergeometric function through a power series and see how to solve the integral to obtain that kind of result explicitly.