To clarify, for those who have not looked at the reference, the integral identity in question is
$$
\int_{\mathbb{R}^d} \frac{d^dq}{(q^2)^{\nu_1} ((k-q)^2)^{\nu_2}}
= \frac{\Gamma(d/2-\nu_1)\Gamma(d/2-\nu_2)\Gamma(\nu_1+\nu_2-d/2)}
{\Gamma(\nu_1)\Gamma(\nu_2)\Gamma(d-\nu_1-\nu_2)}
\pi^{d/2} (k^2)^{d/2-\nu_1-\nu_2} ,
$$
where $q^2$ is standard Euclidean norm squared. I think this is not the standard way of deriving the answer, but the quickest way I see to get it is notice that this integral is a convolution, $$(2\pi)^d \frac{1}{(k^2)^{\nu_1}}*\frac{1}{(k^2)^{\nu_2}},$$ of two distributions and to use Fourier transforms to simplify it.
Basically, you need to take the Fourier transforms of $1/(q^2)^{\nu_1}$ and $1/(q^2)^{\nu_2}$ (that would be from momentum space to position space, in physics terminology), multiply them, and Fourier transform back. Luckily, the Fourier transforms of such distributions are well known. An explicit formula is given, for instance in formula 2.5 of the Fourier Transform Table Appendix of Gelfand & Shilov's Generalized Functions. Volume 1:
\begin{align}
\mathcal{F}[|x|^\lambda]
&= \frac{\Gamma(\lambda/2+d/2)}{\Gamma(-\lambda/2)} 2^{\lambda+d} \pi^{d/2}
|q|^{-\lambda-d} , \\
\mathcal{F}^{-1}[|q|^\nu]
&= \frac{\Gamma(\nu/2+d/2)}{\Gamma(-\nu/2)} 2^\nu \pi^{-d/2}
|x|^{-\nu-d} ,
\end{align}
where $|q| = \sqrt{q^2}$, and with the conventions $\mathcal{F}[\delta(x)] = 1$, $\mathcal{F}[1] = (2\pi)^d\delta(q)$. If $q^2$ is computed using the Lorentzian inner product instead of the Euclidean one, similar formulas can be found, but they are more subtle: see formulas 2.13--16 of the same table in Gelfand & Shilov.
Putting this together with the convolution formula, and the fact that the product of two powers of $|x|$ is a power of $|x|$, the above integral is then
$$
(2\pi)^d \frac{1}{(k^2)^{\nu_1}} * \frac{1}{(k^2)^{\nu_2}}
= \frac{\Gamma(d/2-\nu_1)\Gamma(d/2-\nu_2)\Gamma(\nu_1+\nu_2-d/2)}
{\Gamma(\nu_1)\Gamma(\nu_2)\Gamma(d-\nu_1-\nu_2)}
\pi^{d/2} \frac{1}{(k^2)^{\nu_1+\nu_2 - d/2}} ,
$$
which is the desired form. The arithmetic can easily be checked by hand or with a computer algebra system (as I did it).
As for your last question, which asks for the pole structure of the above expression as a function of $d$, you need only know that $\Gamma(\nu)$ is never zero for real $\nu$ and that it has simple poles at $\nu=0$ and all negative integers, while the functional identity $\Gamma(\nu)\Gamma(1-\nu) = \pi/\sin(\pi \nu)$ tells you what the residues are.
