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Question Number 133964 by mathocean1 last updated on 26/Feb/21

$$calculate$$$$I={\int }_0^1\frac{1}{1+e^x}dx$$

Answered by bobhans last updated on 26/Feb/21

$$\int \frac{{\mathrm{e}}^{\mathrm{x}}+1-{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=\mathrm{x}-\int \frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}$$$$=\mathrm{x}-\ln (1+{\mathrm{e}}^{\mathrm{x}})+\mathrm{c}$$$$$$$$\mathcal{I}={[\mathrm{x}-\ln (1+{\mathrm{e}}^{\mathrm{x}})]}_0^1=1-\ln (1+\mathrm{e})+\ln \left(2\right)$$$$=1+\ln \left(\frac{2}{1+\mathrm{e}}\right)=\ln \left(\frac{2\mathrm{e}}{1+\mathrm{e}}\right)$$

Answered by Olaf last updated on 26/Feb/21

$$\mathrm{I}={\int }_0^1\frac{dx}{1+e^x}$$$$\mathrm{I}={\int }_0^1(\frac{1+e^x}{1+e^x}-\frac{e^x}{1+e^x})dx$$$$\mathrm{I}={[x-\ln (1+e^x)]}_0^1$$$$\mathrm{I}=1-\ln (1+e)+\ln 2$$$$\mathrm{I}=1+\ln \left(\frac{2}{1+e}\right)$$

Answered by mathmax by abdo last updated on 26/Feb/21

$$\mathrm{I}={\int }_0^1\frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{changement}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\mathrm{give}\mathrm{x}=\mathrm{lnt}\Rightarrow$$$$\mathrm{I}={\int }_1^{\mathrm{e}}\frac{\mathrm{dt}}{\mathrm{t}(\mathrm{t}+1)}={\int }_1^{\mathrm{e}}(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}+1})\mathrm{dt}={[\ln \mid \frac{\mathrm{t}}{\mathrm{t}+1}\mid ]}_1^{\mathrm{e}}=\ln \left(\frac{\mathrm{e}}{\mathrm{e}+1}\right)-\ln \left(\frac{1}{2}\right)$$$$=1-\ln (\mathrm{e}+1)+\ln \left(2\right)$$

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