Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then, 
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx 
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right)
\end{align}
The pole of the function at $i$ is of order $n+1$, so the residue is computed by
\begin{align}
\operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\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}(-1)^n\frac{(n+1)(n+2)\cdots(2n+1)}{(z+i)^{2n+1}} \\
&=\frac{(2n+1)!}{i2^{2n+1}(n!)^2}
\end{align}
Hence, $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx = \pi\frac{(2n+1)!}{2^{2n}(n!)^2}$$ The answer provided is $\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\pi$. How do I manipulate my answer to obtain this answer?

Your answer is actually
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}.$$
(Check the derivative.)
Then use
$$ (2n)!=(1 \cdot 3 \cdots (2n-1))\cdot 2^n \cdot (n!)$$ to get
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2^n \cdot (n!)}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)}.$$
