Integral of the function $ (1+|x|^2)^{-k}$
$$
\begin{align}
\int_{\mathbb{R}^n}{\frac{\mathrm{d}x}{\left(1+|x|^2\right)^k}}
&=\omega_{n-1}\int_0^\infty\frac{r^{n-1}\,\mathrm{d}r}{\left(1+r^2\right)^k}\tag1\\
&=\frac{\omega_{n-1}}2\int_0^\infty\frac{r^{\frac n2-1}\,\mathrm{d}r}{\left(1+r\right)^k}\tag2\\
&=\frac{\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\frac{\Gamma\!\left(\frac n2\right)\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}\tag3\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^{\frac n2}\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}}\tag4
\end{align}
$$
Explanation:
$(1)$: convert to polar coordinates
$(2)$: substitute $r\mapsto r^{\frac12}$
$(3)$: apply $(9)$ and the Beta Function
$(4)$: cancel common factors
Formula $(4)$ is finite for $k\gt\frac n2$.
Computation of $\boldsymbol{\omega_{n-1}}$
$$
\begin{align}
1
&=\int_{\mathbb{R}^n} e^{-\pi|x|^2}\mathrm{d}x\tag5\\
&=\omega_{n-1}\int_0^\infty e^{-\pi r^2}r^{n-1}\,\mathrm{d}r\tag6\\
&=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r}r^{\frac n2-1}\,\mathrm{d}r\tag7\\
&=\frac{\omega_{n-1}}2\pi^{-\frac n2}\Gamma\!\left(\frac n2\right)\tag8
\end{align}
$$
Explanation:
$(5)$: integral is the product of $n$ copies of $\int_{-\infty}^\infty e^{-\pi x^2}\mathrm{d}x=1$
$(6)$: convert to polar coordinates
$(7)$: substitute $r\mapsto r^{\frac12}$
$(8)$: apply the Gamma Function
Equation $(8)$ implies $$ \omega_{n-1}=\frac{2\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\tag9 $$