<s>[INST] 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 
 [/INST] Mathematical induction and trigonometric function relations and derivatives should do it.
I'll assume $x/a\in (0,\pi).$
Step: $n = 0$
$$\frac{1}{x^2+a^2}=\frac{\sin{[\mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{1/2}}$$ as
$$\sin{[\mathrm{arccot}{(x/a)}]} = \frac{a}{(a^2+x^2)^{1/2}}. $$
Step: $n \Rightarrow n+1$
Assume
$$\left(\frac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\frac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
it is true, and prove
$$\left(\frac{1}{x^2+a^2}\right)^{(n+1)}=(-1)^{(n+1)}(n+1)!\frac{\sin{[(n+2)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+2)/2}}.$$
We now need to show:
$$\left((-1)^{(n)}n!\frac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}\right)^\prime = (-1)^{(n+1)}(n+1)!\frac{\sin{[(n+2)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+2)/2}}.$$
This is due to:
$$(f/g)^\prime = (f^\prime g - fg^\prime)/g^2$$
and
$$ \sin(x+y) = \sin x\cos y + \sin y \cos x$$
and
$$ \left(\mathrm{arccot}{(x)}\right)^\prime =-\frac{1}{1+x^2}. $$
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