Show that: $\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}$ I was observing this paper and by some experimental chance I got this result, but I can not prove it.

$$\sum_{k=0}^{\infty}\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\frac{\pi}{4}\tag1$$

I tried to factorise $4k^4+12k^3+13k^2+6k+5$ but it is not factorable.
I guess that $$\arctan\left(\frac{6k^2+4k}{4k^4+12k^3+13k^2+6k+5}\right)=\arctan(x)+\arctan(y)$$
Does anyone knows how to prove $(1)?$

Let $ u = \frac{2}{ k^2 + 2k + 1}, v = \frac{2}{  4k^2 + 4k + 1}$. Then,
$$ \arctan (u) - \arctan (v) = \arctan ( \frac{ u - v} { 1 + uv} ) = \arctan ( \frac{6k^2 + 4k } { 4k^4 + 12k^3 + 13k^2 + 6k + 5 }). $$

Then use $\sum_{k=1} \arctan \frac{2}{k^2} = \frac{3 \pi } { 4} $ (2.14) and $ \sum_{k=0} \arctan \frac{2}{(2k+1)^2 } = \frac{ \pi}{2}$ (2.8) as given in the paper, to justify
$\sum_{k=0}  \arctan ( \frac{6k^2 + 4k } { 4k^4 + 12k^3 + 13k^2 + 6k + 5 }) $
$= \sum_{k=0} \arctan \frac{2}{(k+1)^2} - \arctan \frac{2}{(2k+1)^2} $
$ = \sum_{k=1} \arctan\frac{2}{k^2} - \sum_{k=0} \arctan \frac{2}{ (2k+1)^2 } $
$ = \frac{ 3 \pi }{4} - \frac{\pi}{2} $
$= \frac{ \pi}{4} .$
Note: The 2 identities given in the paper can again be proved via telescoping sums.
2.8 is done via 1 telescoping sum ($u=2k, v = 2k+2)$.
2.14 is done via 2 telescoping sums.   
