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I wanted to calculate this one$\displaystyle\int\limits_{0}^{\infty} \dfrac{\sin 2x}{x(\cos x+\cosh x)}\; dx$

I used $\displaystyle 2\sum\limits_{k=1}^{\infty}(-1)^{k-1}\sin kx e^{-kx}=\dfrac{\sin x}{\cos x+\cosh x}$ and after all efforts it reduced to

$\displaystyle \sum\limits_{k=1}^{\infty} (-1)^{k-1} \arctan(2k^2)$ and now this seems quite confusing to me. Please share if anyone has any inspiring solution.

Maybe I've done some mistake while reducing it to a sum,

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    The arctan sum doesn't converge, as the terms don't go to zero. – 2017-02-14

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Hint

You nicely obtained $$\displaystyle\int\limits_{0}^{\infty} \dfrac{\sin 2x}{x(\cos x+\cosh x)}\; dx=\color{red}{2}\sum\limits_{k=1}^{\infty} (-1)^{k-1} \arctan(2k^2)$$ Now, consider $$u_k=(-1)^{k-1} \arctan(2k^2)$$ which gives $$u_{k+1}+u_k=(-1)^k \left(\arctan(2(k+1)^2)-\arctan(2k^2) \right)$$ which simplifies to $$u_{k+1}+u_k=(-1)^k \tan ^{-1}\left(\frac{2(k+1)}{1+4 k^2 (k+1)^2}\right)$$ and, as MyGlasses commented, you face a convergent series.

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    So, the last series you obtained is convergent. Couldn't we evaluate this in a closed form ? Or isn't there an alternate approach the integral. – 2017-02-14
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    @AdityaNarayanSharma. Be sure I tried but I did not get it. – 2017-02-14
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    the integral indeed should posess a closed form solution. i did an almost similar calculation here: http://math.stackexchange.com/questions/2071697/prove-this-bizarre-integral-int-0-infty-sinx-pi-over4-over-x-cosx/2071758#2071758 – 2017-02-14