Find the closed form for $\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}$ 
$$I=\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}\tag1$$

$$\ln\left({1+e^{-x}\over 1-e^{-x}}\right)=2\sum_{n=0}^{\infty}{e^{-(2n+1)x}\over 2n+1}\tag2$$
Sub $(2)$ into $(1)\rightarrow (3)$
$$I=2\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{\infty}e^{-(2n+1)x}\cos{x}dx\tag3$$
Apply integration by parts to $(4)$
Hence
$$I=\int_{0}^{\infty}e^{-(2n+1)x}\cos{x}dx={2n+1\over (2n+1)^2+1}\tag4$$
Apply $(4)$ into $(3)$
Hence
$$I=\sum_{n=0}^{\infty}{2\over (2n+1)^2+1}\tag5$$
Simplify $${2\over (2n+1)^2+1}={2\over 4n^2+4n+2}={1\over n^2+(n+1)^2}$$
Therefore
$$I=\sum_{n=0}^{\infty}{1\over n^2+(n+1)^2}\tag6$$
I am not able to determine the closed form for $(1)$, can anyone please help?

Edit(hint from Marco)

$$I=\int_{0}^{\infty}\cos{x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx={\pi\over 2}\tanh\left({\pi\over 2}\right)\tag1$$

Can anybody prove $(1)$ using another method?

Another approach, just for fun. Using Cantarini's lemma:
$$ \frac{1}{a^2+b^2}=\int_{0}^{+\infty}\frac{\sin(ax)}{a}e^{-bx}\,dx \tag{1}$$
and the fact that $ g(x)=\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1}$ is the Fourier series of a rectangle wave $r(x)$ we have:

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{n^2+(n+1)^2} = 2\sum_{n\geq 1}\frac{1}{(2n+1)^2+1} &=& -1+2\int_{0}^{+\infty}r(x)\,e^{-x}\,dx\\&=&-1+\frac{\pi}{2}\left(\int_{0}^{\pi}e^{-x}\,dx\right)\sum_{k\geq 0}(-1)^k e^{-k\pi}\\&=&-1+\frac{\pi}{2}(1-e^{-\pi})\frac{1}{1+e^{-\pi}}\\&=&\color{red}{-1+\frac{\pi}{2}\,\tanh\left(\frac{\pi}{2}\right)}.\tag{2} \end{eqnarray*}$$


Another creative approach comes from computing the Fourier (cosine) transform of $\frac{\tanh x}{x}$, then prove the wanted result from Fourier inversion. But the Fourier transform of $\frac{\tanh x}{x}$ (a re-scaled $\log\coth$) can be found through the complex version of Frullani's theorem, the fact that $\text{Re}\log(z)=\log\|z\|$ and the Weierstrass products for $\sin$ and $\coth$. So, from Marco to Marco :D

But, wait: this is just the same as applying Poisson's summation formula to the RHS of OP's $(1)$!

