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 
 A: 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
$$ \left(\mathrm{arccot}{(x)}\right)^\prime =-\frac{1}{1+x^2}. $$
A: First Method
Let $$\mathcal{L}\{f(t)\}=\int_0^\infty f(t)\operatorname{e}^{-xt}\operatorname{d}t=F(x)$$ be the Laplace transform of $f(t)$. 
For $f(t)=\sin(at)u(t)$ then we have $$F(x)=\int_0^\infty \sin(at)\operatorname{e}^{-xt}\operatorname{d}t=\frac{a}{x^2+a^2}.$$
Recalling that $\mathcal{L}\{t^n f(t)\}=(-1)^n F^{(n)}(x)$ we have
$$\begin{align}
\frac{(-1)^nF^{(n)}(x)}{a}=\left(\frac{1}{x^2+a^2}\right)^{(n)}&=\frac{(-1)^n\mathcal{L}\{t^n f(t)\}}{a}=\frac{(-1)^n}{a}\int_0^\infty t^n \sin(at)\operatorname{e}^{-xt}\operatorname{d}t\\ 
&=\frac{(-1)^n}{2ia}\left[\int_0^\infty t^n \operatorname{e}^{-(x-ia)t}\operatorname{d}t-\int_0^\infty t^n \operatorname{e}^{-(x+ia)t}\operatorname{d}t\right]\\
&=\frac{(-1)^n}{2ia}\Gamma(n+1)\left[\frac{1}{(x-ia)^{n+1}}-\frac{1}{(x+ia)^{n+1}}\right]
\end{align}
$$ 
using the identity $\sin(at)=\frac{\operatorname{e}^{iat}-\operatorname{e}^{-iat}}{2i}$ and the Gamma Function.
For $x=a\cot{\theta},\,0<\theta<\pi,$ we have
$$
x\pm ia=\frac{a}{\sin\theta}\operatorname{e}^{\pm i\theta}={(x^2+a^2)^{1/2}}\operatorname{e}^{\pm i\theta}
$$
and finally
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=\frac{(-1)^n}{2ia}\Gamma(n+1)\frac{\left[\operatorname{e}^{+i(n+1)\theta}-\operatorname{e}^{-i(n+1)\theta}\right]}{(x^2+a^2)^{\frac{n+1}{2}}}=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
$$
Second Method
Observing that $$\frac{1}{x^2+a^2}=\frac{1}{x+ia}\cdot\frac{1}{x-ia}$$ and using the general Leibniz rule $$
    (f \cdot g)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} g^{(n-k)} $$ with $f(x)=\frac{1}{x+ia}$ and $g(x)=\frac{1}{x-ia}$ we have $$\frac{\operatorname{d}^n}{\operatorname{d}x^n}(x\pm ia)^{-1}=(-1)^n n!(x\pm ia)^{-(n+1)}$$
and then
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}&=\sum_{k=0}^n {n \choose k} (-1)^k k!(x+ ia)^{-(k+1)}(-1)^{n-k} (n-k)!(x- ia)^{-(n-k+1)}\\
&=(-1)^n n!\sum_{k=0}^n (x+ ia)^{-(k+1)}(x- ia)^{-(n-k+1)}.
\end{align}
$$
For $x=a\cot{\theta},\,0<\theta<\pi,$ we have
$$
x\pm ia=\frac{a}{\sin\theta}\operatorname{e}^{\pm i\theta}={(x^2+a^2)^{1/2}}\operatorname{e}^{\pm i\theta}
$$
and then
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}
&=(-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\sum_{k=0}^n \operatorname{e}^{+i(n-2k)\theta}\\
&= (-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\operatorname{e}^{+i n\theta}\frac{1-\operatorname{e}^{-2i\theta(n+1)}}{1-\operatorname{e}^{-2i\theta}}
\end{align}
$$
using the geometric sum $ \sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z} $ with $z=\operatorname{e}^{-2i\theta}$.
Using the Euler's identity $\operatorname{e}^{+i\varphi}-\operatorname{e}^{-i\varphi}=2i\sin\varphi$ and multiplying and dividing by $a\operatorname{e}^{i\theta}$ we obtain
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}
&= (-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\frac{a\operatorname{e}^{i\theta}}{a\operatorname{e}^{i\theta}}\frac{\operatorname{e}^{i\theta(n+1)}-\operatorname{e}^{-i\theta(n+1)}}{\operatorname{e}^{i\theta}-\operatorname{e}^{-i\theta}}\\
&=(-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\frac{a}{\sin\theta}\frac{1}{a}\sin((n+1)\theta)\\
&=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
\end{align}
$$
Third Method
Let be $$f(x)=\frac{1}{x^2+a^2}=\frac{1}{a^2}\frac{1}{1+t^2}=\frac{1}{a^2}\psi(t).$$ Observe that $$\frac{\operatorname{d}^n f(x)}{\operatorname{d}x^n}=\frac{1}{a^{n+2}}\frac{\operatorname{d}^{n+1} \psi(t)}{\operatorname{d}t^{n+1}}$$ where $\psi(t)=\arctan(t)$.
Putting $\sin\theta=\frac{1}{\sqrt{1+t^2}}$ the $n$th-derivative of $\psi(t)$ is
$$
\psi^{(n)}(t)=(-1)^{n-1}(n-1)!\sin^n\theta\sin(n\theta)\tag 1
$$
The existence of the derivatives follows from the analyticity of $\arctan t$ on the real line. The proof of formula (1) is by mathematical induction. Clearly, the (1) is true
for $n = 1$. Suppose the (1)is true for $n = k$; that is, suppose 
$$
\psi^{(k)}(t)=(-1)^{k-1}(k-1)!\sin^k\theta\sin(k\theta)\tag 2
$$
We will show that the (1) is true for $n = k + 1$ whenever it is true for $n = k$.
Diﬀerentiating both sides of (2) with respect to $t$ and noting that $\frac{\operatorname{d} \theta}{\operatorname{d}t}=-\sin^2\theta$ gives
$$\frac{\operatorname{d}\psi^{(k)}(t)}{\operatorname{d}t}=(-1)^{k}k!\sin^{k+1}\theta[\cos\theta\sin(k\theta)+\cos(k\theta)\sin\theta]=(-1)^{k}k!\sin^{k+1}\theta\sin((k+1)\theta)$$
that is $$\psi^{(k+1)}(t)= (-1)^{k}k!\sin^{k+1}\theta\sin((k+1)\theta)$$ so the (1) is true for any $n\ge 1$.
Thus we have
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=\frac{1}{a^{n+2}}\psi^{(n+1)}(t)=\frac{1}{a^{n+2}}(-1)^{n}n!\sin^{n+1}\theta\sin((n+1)\theta)
$$
and observing that $\frac{\sin\theta}{a}=\frac{1}{a\sqrt{1+t^2}}=\frac{1}{(x^2+a^2)^{1/2}}$ finally we obtain
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
$$
A: 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.
