$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.

According to Maple,
$$ \frac{\pi}{\sqrt{3}} \tanh \left(\frac{\pi \sqrt{3}}{2}\right) - 1 $$
It looks like all the sums $\sum_{n=1}^\infty \frac{1}{n^2+tn+1}$ for integers $t\ne -2 $ have closed forms.
EDIT: Here are a few examples:
$$ \eqalign{\sum_{n=1}^{\infty } \frac{1}{ {n}^{2}+n+1} & =\frac{\pi \tanh(\pi \sqrt{3}/2)}{\sqrt{3}}-1\cr
\sum_{n=1}^\infty \frac{1}{n^2+n+1} &= \frac{\pi^2}{6}-1\cr
\sum_{n=1}^\infty \frac{1}{n^2+3n+1} &= \frac{\pi \tan(\pi \sqrt{5}/2)}{\sqrt{5}}\cr
\sum_{n=1}^\infty \frac{1}{n^2+4n+1} &= -\frac{\pi \cot(\pi \sqrt{3})}{2\sqrt{3}}-\frac{1}{3}\cr
\sum_{n=1}^\infty \frac{1}{n^2+5n+1} &= \frac{\pi \tan(\pi \sqrt{21}/2)}{\sqrt{21}}-\frac{7}{15}\cr
\sum_{n=1}^\infty \frac{1}{n^2+6n+1} &= -\frac{\pi\cot(2 \sqrt{2}\pi)}{4\sqrt{2}}-\frac{61}{112}}$$
