Integration by Substitution in $\mathbb{R}^{n}$ We are given, as a hint, that the $(n - 1)$-dimensional volume of the unit ball in $\mathbb{R}^{n}$ is $2\pi^{\frac{n}{2}} / \Gamma(\frac{n}{2})$. 
We are also given the following integral: $\int_{0}^{\infty} \frac{s^{\alpha}}{(1+s^{2})^{\beta}} \text{d}s = \Gamma(\frac{1+\alpha}{2})\Gamma(\beta - \frac{1+\alpha}{2}) / 2\Gamma(\beta)$, if $\alpha, \beta > 0$, and $2\beta - \alpha - 1 > 0$.
During the related problem, I have ended up with the following integral:
$\int_{\mathbb{R^{n}}} \frac{|x|^2}{(1 + |x|^2)^n} \text{d}x$
Clearly, I would like to substitute $s = |x|$ to use the integral provided to us above. However, I am unsure what the d$x$ in my integral would become, in the $n$-dimensional case.
If I recall correctly, in the $\mathbb{R}^3$ case, we would do the following:
$\int_{\mathbb{R^{3}}} \frac{|x|^2}{(1 + |x|^2)^3} \text{d}x = \int_{0}^{2\pi} \text{d}\theta \int_{0}^{\pi} \text{sin} \phi  \ \text{d}\phi \int_{0}^{\infty} \frac{s^{2}}{(1+s^{2})^{3}} s^{2} \text{d}s = 4\pi \int_{0}^{\infty} \frac{s^{4}}{(1+s^{2})^{3}} \text{d}s$.
Here, the d$x$ is replaced by $\sin \phi \ s \ \text{d}s \ \text{d}\phi \ \text{d} \theta$. What would be the similar transformation for $n$ dimensions? The $\phi$ and $\theta$ parts of this integral just became the factor of $4\pi$, which matches the unit ball equation we were are given. The part that worries me is the extra $s^2$ factor that appears. 
 A: $2\pi^{\frac n 2}/\Gamma(\frac n 2)$ is not the volume of a (n-1)-dimensional unit ball, but the volume of a (n-1)-dimensional unit sphere (which is the boundary of a n-dimensional ball). You can check that for low $n$
\begin{array}{c|cc}n & V_{n-1}(r) & S_{n-1}(r) \\\hline 2 & 2r & 2\pi r \\ 3 & \pi r^2 & 4\pi r^2\end{array}
and the formula you were given gives you $S_{n-1}(1)$, not $V_{n-1}(1)$.
In $n$ dimension you can write
$$ {\rm d}x = s^{n-1}{\rm d}s\, {\rm d}\Omega $$
where $s=|x|$ and ${\rm d}\Omega$ encompasses all polar angles and their trigonometric functions. You'll have then
$$ \int_{\mathbb R^n} \frac{|x|^2}{(1+|x|^2)^m} {\rm d}x = \int_{\mathbb S^{n-1}} {\rm d}\Omega \int_0^\infty \frac{s^{n+1}ds}{(1+s^2)^m}$$
Integral $\int_{\mathbb S^{n-1}} {\rm d}\Omega $ is equal to the area of (n-1)-dimensional unit sphere, so you have
$$ \int_{\mathbb R^n} \frac{|x|^2}{(1+|x|^2)^m} {\rm d}x = \frac{2\pi^{\frac n 2}}{\Gamma(\frac n 2)} \int_0^\infty \frac{s^{n+1}ds}{(1+s^2)^m}$$
A: If it a unit ball in $\mathbb{R}^n$, then it is uniform between each of its quadrants (but this name only applies when there are four). We can notice that for an $n$ dimensional Cartesian plane there are $2^n$ identical areas ($4$ for $2D$, $8$ for $3D$ etc.) so you could use this to get rid of the need for modulus signs
