$$\begin {aligned} &\int _{0}^{\infty }\arctan \frac {\sinh x}{1+\cosh 2x}dx\\ &=\int _{0}^{\infty }\arctan \frac {2\sinh x-\sinh x}{1+2\sinh ^{2}x}dx\\ &=\int _{0}^{\infty }\left (\arctan 2\sinh x-\arctan \sinh x\right )dx\\ &=\int _{0}^{\infty }\int _{1}^{2}\frac {\sinh x}{1+t^{2}\sinh ^{2}x}dtdx\\ &=\int _{1}^{2}\int _{0}^{\infty }\frac {\sinh x}{t^{2}\cosh ^{2}x-t^{2}+1}dxdt\\ &=\int _{1}^{2}\frac {1}{2t\sqrt {t^{2}-1}}\left .\ln \frac {t\cosh x-\sqrt {t^{2}-1}}{t\cosh x+\sqrt {t^{2}-1}}\right |_{x=0}^\infty dt\\ &=\int _{1}^{2}\frac {\ln \left (t+\sqrt {t^{2}-1}\right )}{t\sqrt {t^{2}-1}}dt\\ &=\int _{\frac {1}2}^{1}\ln \left (\frac {1+\sqrt {1-u^{2}}}u \right )\frac {du}{\sqrt {1-u^{2}}}\\ &=\int _{\frac {\ti}{12}}^{\frac {\ti }4}\ln \left (\frac {1+\cos \theta }{\sin \theta }\right )d\theta\\ &= \int _{\frac {\ti }{12}}^{\frac {\ti }4}\ln \cot \frac {\theta }2d\theta \\ &=\int _{\frac {\ti }{12}}^{\frac {\ti }4}\sum _{n=1}^\infty \frac {2\cos (2n-1)\theta }{2n-1}d\theta \\ &=2\sum _{n=1}^\infty \frac {1}{(2n-1)^{2}}\left [\sin (2n-1)\frac {\ti }4-\sin (2n-1)\frac {\ti }{12}\right ]\\ &=2\left (\frac {1}2\sum _{n=1}^\infty \frac {(-1)^{n-1}}{(2n-1)^{2}}-\frac {3}2\sum _{n=1}^\infty \frac {(-1)^{n-1}}{(6n-3)^{2}}\right )\\ &=\beta (2)-\frac {1}3\beta (2)\\ &=\textcolor {blue}{\frac {2}3\beta (2)}. \end {aligned}$$