compute integral for $n\ge 3$ I was trying to prove Sobolev inequality and estimate the value of the constant for dimensions greater (or equal) to 3 and these integrals came up
$$I_\pm=\int_0^\infty \frac{r^{n\pm1}}{(b^2+r^2)^n} dr$$
I tried substitutions $r=b\tan t$, $r=b\sinh t$ and even Cauchy formula for integrals... but I didn't get any results. Any hint?
 A: If you consider the problem of the antiderivative $$J=\int \frac{r^{a}}{(b^2+r^2)^n}\, dr$$ the result is given in terms of hypergeometric function
$$J=\frac{r^{a+1}  \, _2F_1\left(1,\frac{1}{2} (a-2
   n+3);\frac{a+3}{2};-\frac{r^2}{b^2}\right)}{(a+1) b^2\left(b^2+r^2\right)^{n-1}}$$ making 
$$I_a=\int_0^\infty \frac{r^{a}}{(b^2+r^2)^n}\, dr=\frac 12\, b^{(a+1-2 n)}\,\,\frac{\Gamma \left(\frac{a+1}{2}\right)\, \Gamma
   \left(n-\frac{a+1}{2}\right)}{\Gamma (n)}$$ provided
$$\Re(b)\neq 0\land \Re(a-2 n)<-1\land (a+1) \arg \left(\frac{1}{b^2}\right)\leq 2
   \pi \land \Re(a)>-1$$
Edit
Otherwise, use
$$r=b \tan(x) \qquad dr=\frac{dx} {\cos^2(x)}\qquad 1+\tan^2(x)=\frac{1} {\cos^2(x)}$$ to make
$$I_a=\int_0^\infty \frac{r^{a}}{(b^2+r^2)^n}\, dr=b^{(a+1-2n)} \int_0^{\frac \pi 2} \sin^a(x) \cos^{(2n-a-2)}(x) \,dx$$ which is a classical one (since, morover, in your case $a$ is an integer).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
I_{\pm} & \equiv \int_{0}^{\infty}{r^{n \pm 1} \over \pars{b^{2} + r^{2}}^{n}} \,\dd r
\,\,\,\stackrel{{\large{r/\verts{b}}}\ \mapsto\ r}{=}\,\,\,
{1 \over \verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}{r^{n \pm 1} \over
\pars{r^{2} + 1}^{n}}\,\dd r
\\[5mm] \stackrel{r^{2}\ \mapsto\ r}{=}\,\,\,&
{1 \over 2\verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}{r^{n/2 \pm 1/2 - 1/2} \over
\pars{r + 1}^{n}}\,\dd r
\\ = &\
{1 \over 2\verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}r^{n/2 \pm 1/2 - 1/2}\
\overbrace{\bracks{{1 \over \Gamma\pars{n}}\int_{0}^{\infty}t^{n - 1}\expo{-\pars{r + 1}t}\,\dd t}}^{\ds{1 \over \pars{r + 1}^{n}}}\
\,\dd r\quad\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] = &\
{1 \over 2\verts{b}^{n - 1 \mp 1}\,\Gamma\pars{n}}
\int_{0}^{\infty}t^{n - 1}\expo{-t}\
\overbrace{\int_{0}^{\infty}r^{n/2 \pm 1/2 - 1/2}\expo{-tr}\,\dd r}
^{\ds{\Gamma\pars{n/2 \pm 1/2 + 1/2}/t^{n/2 \pm 1/2 + 1/2}}}\,\dd t
\\[5mm] = &\
{1 \over 2\verts{b}^{n - 1 \mp 1}}\,{\Gamma\pars{n/2 \pm 1/2 + 1/2} \over \Gamma\pars{n}}
\int_{0}^{\infty}t^{n/2 \mp 1/2 - 3/2}\expo{-t}\dd t
\\[5mm] = &\
{1 \over 2\verts{b}^{n - 1 \mp 1}}\,{\Gamma\pars{n/2 \pm 1/2 + 1/2} \over \Gamma\pars{n}}\,\,\Gamma\pars{{n \over 2} \mp {1 \over 2} - {1 \over 2}}
\\[5mm] = &\
\bbx{{\mrm{B}\pars{n/2 \pm 1/2 + 1/2,n/2 \mp 1/2 - 1/2} \over
2\verts{b}^{n - 1 \mp 1}}}
\qquad\pars{~\mrm{B}:\ Beta\ Function~}
\end{align}
