\begin{align} \sum_{n=1}^{\infty}\frac{n}{n^4+4} &=\sum_{n=1}^{\infty}\frac{n}{(n^2+2n+2)(n^2-2n+2)}\\ &=\frac{1}{4}\times\sum_{n=1}^{\infty}\Big(\frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2}\Big)\\ &=\frac{1}{4}\times\sum_{n=1}^{\infty}\Big(\frac{1}{n^2-2n+2}-\frac{1}{n^+2n+2}\Big)\\ &=\frac{1}{4}\times\sum_{n=1}^{\infty}\Big(\frac{1}{(n-1)^2+1}-\frac{1}{(n+1)^2+1}\Big)\\ &=\frac{1}{4}\times\Big(\frac{1}{0^2+1}+\frac{1}{1^2+1}\Big)\\ &=\boxed{\frac{3}{8}} \end{align} Note: the first step utlizes the Sophie Germain's Identity (see %%HREF%%3863%%).