<s>[INST] Find the infinite sum of the series $\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}$ WolframAlpha outputs the answer as
$$-\frac14 \pi\left(-\coth(\pi) + \pi \operatorname{csch}^2(\pi)\right)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
 [/INST] First, we note that $$\frac{n^2}{(1+n^2)^2}=\color{blue}{\frac1{1+n^2}}-\color{orange}
{\frac{1}{(1+n^2)^2}}.$$
So $$\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}-\color{orange}{\sum_{n=1}^\infty
\frac{1}{(1+n^2)^2}}.$$
Using the identities $$\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}=\frac12\pi\coth(\pi)-\frac12$$
from Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$ and 
$$\color{orange}{\sum_{n=1}^\infty
\frac{1}{(1+n^2)^2}}=-\frac{1}{2}+\frac{1}{4} \pi  \coth (\pi )+\frac{1}{4} \pi ^2 \text{csch}^2(\pi )$$ from How to calculate $\sum_{k=1}^{\infty}\frac{1}{(a^2+k^2)^2}$ after calculating $\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}$ using Parseval identity?, we reach the final result $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\frac{1}{4} \pi  \coth (\pi )-\frac{1}{4} \pi ^2 \text{csch}^2(\pi )}$$
</s>