How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=\dfrac{(-1)^nn!}{2ai}\left(\dfrac{1}{(x-ai)^{n+1}}-\dfrac{1}{(x+ai)^{n+1}}\right)$$
so
let$$x=a\cot{\theta},0<\theta<\pi,$$
then
$$x\pm ai=a(\cos{\theta}\pm i\sin{\theta})/\sin{\theta}$$
so
$$\dfrac{1}{(x\pm ai)^{n+1}}=\dfrac{\sin^{n+1}{\theta}}{a^{n+1}}[\cos{(n+1)\theta}\mp i\sin{(n+1)\theta}]$$
so$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
Question:
Have other methods?
Because this is important reslut,so I think this have other methods?  Thank you 

First, you can change variables to reduce to the case $a=1$.  Second, the integral is the arctangent, so you are asking for the $(n+1)$th derivative of $\arctan(x)$.  Maple says:
$$
\left(\frac{d}{dx}\right)^n\arctan x =
\frac{1}{2}\,{2}^{n}
G^{1, 3}_{3, 3}\left({x}^{2}\, \Big\vert\,^{0, 0, 1/2}_{0, (n-1)/2, n/2}\right)
{x}^{1-n}
$$
in terms of the Meijer G-function.
