Evaluating a complex contour I need to show the following result:
$$
\int_{-\infty}^\infty \frac{1}{(1+x^2)^{n+1}}dx\, = \frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2\cdot 4\cdot\ldots\cdot(2n)}\pi 
$$
With n=1,2,3,...
This function has a pole at i and -i.
I've tried a semicircle in the upperhalf of the plain, but the residue then goes to infinity. I've also tried a rectangle in the upperhalf that stays beneath i, but all 3 sides that do not include the integral we're looking for go to zero because of the R in the denominator.
Anyone with tips?

if x=tany, $dx=sec^2ydy$  so y varies from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
$\int 1 / (1+x^{2})^{n+1}dx = \int \frac{sec^2ydy}{(sec^2y)^{n+1}} dy= \int cos^{2n}y dy = I_{2n}(say)$
Here is how to derive the reduction formula 
So, $I_m=\frac{cos^{m-1}xsinx}{m}+\frac{m-1}{m}I_{m-2}$
$I_{2n}=\frac{cos^{2n-1}xsinx}{2n}+\frac{2n-1}{2n}I_{2n-2}$
Now if take definite integral of $I_{2n}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$,
$\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n}y dy =[\frac{cos^{2n-1}xsinx}{2n}]_{-\frac{\pi}{2}}^\frac{\pi}{2}+\frac{2n-1}{2n}\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n-2}y dy=\frac{2n-1}{2n}\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n-2}y dy$ (as the first integral is 0)
$I_2= \int cos^2y dy=\int \frac{1+cos2y}{2} dy=\frac{y}{2}+\frac{sin2y}{2}+C$ where C is the indeterminate constant of indefinite integral.
The  definite integral of $I_{2}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ will be $\pi$
So, the  definite integral of $I_{2n}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ will be $\frac{(2n-1)(2n-3)...3.1}{2n(2n-2)...4.2} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^2y dy$
$=\frac{(2n-1)(2n-3)...3.1}{2n(2n-2)...4.2} \pi$
