<s>[INST] show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$ show that:
$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$
where $n=0,1,2,3,\ldots$.
is there any help?
thanks for all
 [/INST] It is easier to find the integral in a more general form below:
$$I_{n}(a)=\int_{-\infty}^{+\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}} \quad \text {, where } n=0,1,2,3, \cdots \textrm{ and }a>0.$$
We first start with the easiest one $I_0(a)$, $$\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\frac{1}{\sqrt{a}} \left[ \tan ^{-1} \frac{x}{\sqrt{a}}\right]_{-\infty}^{\infty}=\frac{\pi}{\sqrt{a}}
$$
In order to arrive at the power of (n+1), I plan to differentiate $I_{0}(a)$ w.r.t $a$  by $n$ times. $$
\frac{d^{n}}{da^{n}}\int_{-\infty}^{+\infty} \frac{d x}{x^{2}+a}=\pi \frac{d^{n}}{da^{n}}\left(a^{-\frac{1}{2}}\right)
$$
$$
\int_{-\infty}^{\infty} \frac{(-1)^{n} \cdot n !}{\left(x^{2}+a\right)^{n+1}} d x=\pi\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-1}{2}\right) a^{-\frac{2 n+1}{2}}
$$
$$
\boxed{\int_{-\infty}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n+1}}=\frac{(2n-1)!!\pi}{2^{n} n !}a^{-\frac{2 n+1}{2}}}
$$
Putting $a=1$ yields the result $$
\int_{-\infty}^{\infty} \frac{dx }{\left(x^{2}+1\right)^{n+1}}=\frac{(2 n-1) ! ! \pi}{2^{n} n !} \cdot \frac{2\cdot4 \cdot 6 \cdots(2 n)}{2 \cdot 4 \cdot 6 \cdots(2 n)}= \frac{(2 n) ! \pi}{2^{2 n}(n !)^{2}}. \quad \blacksquare
$$
</s>