This question already has an answer here:
The identity
$$ \sum_{n\geq 0}\frac{1}{n^2+\alpha^2} = \frac{1}{2\alpha^2}+\frac{\pi}{2\alpha}\cdot\coth(\pi\alpha) \tag{1}$$
that holds for any $\alpha>0$, can be proved through the Weierstrass product for the sine function and the properties of the logarithmic derivative, just like here. For instance, by considering that
$$ \lim_{\alpha\to 0^+}\left(-\frac{1}{2\alpha^2}+\frac{\pi}{2\alpha}\cdot\coth(\pi\alpha)\right) = \frac{\pi^2}{6}\tag{2}$$
we may also easily compute $\zeta(2)$. Another approach for proving $(1)$ relies on:
$$ \int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-x}\,dx = \frac{1}{n^2+1}\tag{3}$$
and the fact that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of a sawtooth wave, giving:
$$ \sum_{n\geq 1}\frac{1}{n^2+1} = \frac{e^{2\pi}}{e^{2\pi}-1}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx\tag{4} $$
from which:
$$ \color{red}{\sum_{n\geq 0}\frac{1}{n^2+1}} = \frac{1+\pi\coth(\pi)}{2} = \color{red}{\frac{1}{2}+\frac{\pi}{2}\cdot\frac{e^{2\pi}+1}{e^{2\pi}-1}} \tag{5}$$
easily follows.