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Question Number 164795 by HongKing last updated on 22/Jan/22

Answered by puissant last updated on 22/Jan/22

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(\frac{1}{sinhx}-\frac{1}{xcoshx})dx$$$$={\int }_0^{\mathrm{\infty }}(\frac{1}{\frac{e^x-e^{-x}}{2}}-\frac{1}{x\frac{e^x+e^{-x}}{2}})dx$$$$={\int }_0^{\mathrm{\infty }}(\frac{2}{e^x(1-e^{-2x})}-\frac{2}{{xe}^x(1+e^{-2x})})dx$$$$={\int }_0^{\mathrm{\infty }}(\frac{2e^{-x}}{1-e^{-2x}}-\frac{2e^{-x}}{x(1+e^{-2x})})dx..$$

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