Evaluating $\int_0^{\infty}  {x^2 \over (x^2+A)^n} dx$ for $A \gt 0$ I'm having trouble with the following integral
$$\int_0^{\infty}  {x^2 \over (x^2+A)^n} \ dx \ \ \text{for} \ \  A \gt 0 $$
EDIT:
I had previously thought that the answer to the integral above was:
$$ {2\over(n-1)(n-2)(n-3)A^{n-3}}$$
But actually this is the answer to:
$$\int_0^{\infty}  {x^5 \over (x^2+A)^n} \ dx \ \ \text{for} \ \  A \gt 0 $$
which is much easier to integrate (this is one book misprint I'll never forget). BD answer below made me see that (he tested it numerically). 
Still, since I spent 4 days trying to solve the wrong integral, it is a relief to see a answer to that and to see how complicated it is. This is a sneaky integral right there, seems to be easy until you get down to business. Too bad I have to choose one answer, there are many good ones.
 A: Let $x=\sqrt{A}\,t$. Then $dx=\sqrt{A}\,dt$ and $x^2=At^2$. After we do the substitution, we end up wanting something like
$$\int_0^\infty \frac{A^{3/2}}{A^n}\frac{t^2\,dt}{(1+t^2)^n}.$$ 
There is no sense in carrying the constant around. Note also that 
$$\frac{t^2}{(1+t^2)^n}=\frac{1+t^2}{(1+t^2)^n}-\frac{1}{(1+t^2)^{n}}=\frac{1}{(1+t^2)^{n-1}}-\frac{1}{(1+t^2)^n}.$$ 
So we will be through if we can handle 
$$\int_0^\infty \frac{dt}{(1+t^2)^m}.$$
There are various ways to do it. For example, let $t=\tan\theta$. Then $dt=\sec^2\theta\,d\theta$, and we arrive at
$$\int_0^{\pi/2} \cos^{2m-2} \theta \,d\theta.$$
This can be done by using a Reduction Formula that we get by integrating by parts.
Indeed we could have made the trigonometric substitution already at the $\frac{t^2}{(1+t^2)^n}$ stage. We can also develop a reduction formula directly, without going through the trigonometric substitution. 
A: Make the change of variables $x=\sqrt{A \frac{u}{1-u}}$ for $0<u<1$. This reduces the integral to the Euler beta-integral:
$$
   \int_0^\infty \frac{x^2}{(x^2+A)^n}\mathrm{d}x = \frac{a^{3-2n}}{2}\int_0^1 u^{1/2} (1-u)^{n-5/2} \mathrm{d}u = \frac{a^{3-2n}}{2} \operatorname{B}\left(\frac{3}{2},n-\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4} a^{3-2n} \frac{\Gamma\left(n-\frac{3}{2}\right)}{\Gamma(n)}
$$ 
where $\operatorname{B}(a,b)$ is the Beta-function.
A: This is intended to be a comment to André's answer (but I haven't got enough reputation!).
Speaking about $\displaystyle I_m = \int \frac1{(1+t^2)^m}dt$, I remember it was the hardest case to consider when integrating rational functions. 
On that occasion, I learned about a particular recursive formula, that I found useful: $$\displaystyle I_m = \frac1{2(m-1)}\left[(2m-3)I_{m-1} + \frac{t}{(1+t^2)^{m-1}}\right]$$
It goes under the name of Ostrogradsky and you can also find it in the Wiki page linked by André.
A: Residue Theory
$$I=\int_0^{\infty}  {x^2 \over (x^2+A)^n} \ dx \hspace{10mm}  \text{suppose  :  } A=a^2 $$
$$ I= \frac{1}{2} \int_{-\infty}^{\infty}  {x^2 \over (x^2+a^2)^n} \ dx=\frac{1}{2}\oint_{c^+} \frac{z^2}{(z^2+a^2)}dz =\frac{1}{2} \left[ 2 \pi i \sum_{i=1}^n R_i^+    \right]=\frac{1}{2}(2 \pi i. R_1)$$
$$  c^+ : \text{ Upper  half plane of complex plane (uhp)} $$
$$  c^- : \text{ Lower  half plane of complex plane (lhp)} $$ 
$$ R_i^+ \rightarrow \text{ Residue of function in singularity  ponits ( that   placed in uhp)}$$
$$ \text{Singularites : } z^2+a^2=0 \rightarrow  \begin{cases} z_1=ia  &\mbox{Acceptable (uhp)}   \\
z_2=-ia  & \mbox{Ineligible (lhp) } \end{cases} \text{poles of order of n}$$
$$ R_1=Residue(\frac{z^2}{(z^2+a^2)},z=ia,\text{pole of order of n}) $$
$$ R_1=\frac{1}{(n-1)!}\lim_{z \rightarrow ia } \left[ \frac{d^{n-1}}{dz^{n-1}}(\frac{(z-ia)^n z^2}{(z-ia)^n (z+ia)^n }) \right]$$
$$ R_1=\frac{1}{(n-1)!}\lim_{z \rightarrow ia } \left[ \frac{d^{n-1}}{dz^{n-1}}(\frac{ z^2}{ (z+ia)^n }) \right] $$
$$ \frac{d^{n}}{dz^{n}}(\frac{ z^2}{ (z+ia)^n })= \frac{ \left[ 2 (iaz)+(n-1) a^2 \right](-1)^{n+1}(n)!}{(z+ia)^{n+2}} $$
