Evaluate integral in terms of the hypergeometric function Mathematica's Integrate() gives the following result:
$$
\int \frac{1}{x^{b}}\left[1-(d/x)^b\right]^{N-1}\,\mathrm  dx= -\frac{x^{1-b}{_2F_1}\left(\frac{b-1}{b}, 1-N, \frac{b-1}{b}+1; (d/x)^b \right)}{b-1}.
$$
I'm not too familiar with the properties of the hypergeometric function (many of which are found at https://dlmf.nist.gov/15), and the steps taken to reach this result are not clear to me. I would appreciate any insight and advice on this particular integral!
 A: I am going to substitute $d\mapsto\delta$ as not to confuse with differential in integral. We write your integral as
$$
I=
\delta^{-b}\int (\delta/x)^b(1-(\delta/x)^b)^{N-1}\,\mathrm  dx.
$$
Upon substituting $u=(\delta/x)^b$, $\mathrm dx=-\frac{1}{b}\delta u^{-\frac{1}{b}-1}\mathrm du$:
$$
I=
-\frac{1}{b}\delta^{1-b}\int u^{-\frac{1}{b}}(1-u)^{N-1}\,\mathrm  du.
$$
Calling on the fundamental theorem of calculus permits us to subsequently write
$$
I=
-\frac{1}{b}\delta^{1-b}\int_0^u z^{-\frac{1}{b}}(1-z)^{N-1}\,\mathrm  dz.
$$
After another substitution of $t=z/u$, $\mathrm dz=u\mathrm dt$:
$$
I=
-\frac{\delta^{1-b}u^{\frac{b-1}{b}}}{b-1}
\underbrace{\frac{b-1}{b}
\int_0^1 \frac{t^{\frac{b-1}{b}-1}(1-t)^{(1+\frac{b-1}{b})-(\frac{b-1}{b})-1}}{(1-ut)^{1-N}}\,\mathrm  dt
}_{{_2F_1}(1-N,1-\frac{1}{b};2-\frac{1}{b};u)},
$$
where the hypergeometric term comes from inspection of DLMF 15.6.1. Reintroducing $u$ and simplifying then gives us
$$
I=
-x^{1-b}
\frac{{_2F_1}\left({1-N,1-\frac{1}{b}\atop 2-\frac{1}{b}};(\delta/x)^b\right)}{b-1},
$$
which is the desired result.
Note:
The $2-N$ parameter the hypergeometric function presented by the OP was a typo and should be $1-N$.
