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 $$ \omega_{n-1}=\frac{2\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\tag9 $$