Taking the semicircular contour $C_R:=\left(\gamma_R:=\{z\in\Bbb C\;:\;|z|=R\,\,,\,Im(z)\geq 0\}\right)\cup [-R,R]\,\,,\,R>1$
the domain within this path contains one pole of the function, and:
$f(z)=\frac{1}{(1+z^2)^{n+1}}\Longrightarrow Rez_{z=i}(f)=\frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left[(z-i)^{n+1}f(z)\right]=$ $=\frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left[\frac{1}{(z+i)^{n+1}}\right]=\frac{1}{n!}\lim_{z\to i}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{(z+i)^{2n+1}}=$ $=\frac{1}{n!}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{(2i)^{2n+1}}=\frac{1}{n!}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{2^{2n+1}(-1)^ni}=$ $=\binom{2n}{n}\frac{1}{2^{2n+1}i}$
We thus get by Cauchy's Integral Theorem that $\oint_{C_R} f(z)dz=\binom{2n}{n}\frac{\pi}{2^{2n}}\,\,\,\,\,(**)$
But on $\,\gamma_R\,$ we have $\,z=Re^{it}\,$ , so $\left|\int_{\gamma_R}f(z)dz\right|\leq\max_{z\in\gamma_R}\left|\frac{1}{(1+R^2e^{2it})^{n+1}}\right|\pi R\leq\frac{\pi R}{(1-R^2)^{n+1}}\xrightarrow [R\to\infty]{} 0$
So taking the limit when $\,R\to\infty\,$ in (**) we get $\binom{2n}{n}\frac{\pi}{2^{2n}}=\lim_{R\to\infty}\oint_{C_R} f(z)dz=\int_{-\infty}^\infty\frac{1}{(1+x^2)^{n+1}}dx$