calculus - Find the closed form for $int_{0}^{infty}cos{x}lnleft({1+e^{-x}over 1-e^{-x}}right)dx=sum_{n=0}^{infty}{1over n^2+(n+1)^2}$

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

$$\ln\left({1+e^{-x}\over 1-e^{-x}}\right)=2\sum_{n=0}^{\infty}{e^{-(2n+1)x}\over 2n+1}\tag2$$

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?

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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?

Answer

Denoting
$$F(a)=\sum_{n=1}^\infty \frac{1}{n^2+a^2}=\frac{\pi a\coth\pi a-1}{2a^2}$$
(see, for example, here), we get
$$I=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2+1}=2\left(\sum_{n=1}^\infty\frac{1}{n^2+1}-\sum_{n=1}^\infty\frac{1}{(2n)^2+1}\right)=2F(1)-\frac12 F\Bigl(\frac{1}{2}\Bigr).$$