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Question Number 205750 by MetaLahor1999 last updated on 29/Mar/24

$${\int }_0^{+\mathrm{\infty }}\frac{1}{1+e^{2x}}dx=?$$

Commented by mokys last updated on 29/Mar/24

$$=-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{-2e^{-2x}}{1+e^{-2x}}dx=\frac{1}{2}{\int }_0^1\frac{du}{1+u}$$$$$$$$=\frac{1}{2}\left(ln2\right)$$$$$$$$AldolaimyMohammad$$

Commented by MetaLahor1999 last updated on 29/Mar/24

$$thanku$$

Commented by mokys last updated on 29/Mar/24

$$welcome$$

Answered by lepuissantcedricjunior last updated on 29/Mar/24

$${\int }_0^{\mathrm{\infty }}\frac{d\mathit{x}}{1+{\mathit{e}}^{2\mathit{x}}}={\int }_0^{\mathrm{\infty }}\frac{{\mathit{e}}^{-2\mathit{x}}}{1+{\mathit{e}}^{-2\mathit{x}}}\mathit{d}\mathit{x}=-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{-2{\mathit{e}}^{-2\mathit{x}}}{1+{\mathit{e}}^{-2\mathit{x}}}\mathit{d}\mathit{x}$$$$=-\frac{1}{2}{\left[\mathit{l}\mathit{n}(1+{\mathit{e}}^{-2\mathit{x}})\right]}_0^{\mathrm{\infty }}$$$$=\mathit{l}\mathit{n}\sqrt{2}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{x}}{1+{\mathit{e}}^{2\mathit{x}}}=\mathit{l}\mathit{n}\left(\sqrt{2}\right)$$$$=====================$$$$.......lepuissant\mathit{D}r.......................$$

Answered by mathzup last updated on 31/Mar/24

$$I={\int }_0^{\mathrm{\infty }}\frac{dx}{1+e^{2x}}{=}_{e^x=t}{\int }_1^{\mathrm{\infty }}\frac{dt}{t(1+t^2)}$$$$={\int }_1^{\mathrm{\infty }}(\frac{1}{t}-\frac{t}{1+t^2})dt$$$$={[lnt-\frac{1}{2}ln(1+t^2)]}_1^{\mathrm{\infty }}={\left[ln\left(\frac{t}{\sqrt{1+t^2}}\right)\right]}_1^{\mathrm{\infty }}$$$$=0-ln\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}ln\left(2\right)$$

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