Integration Questions

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

$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}=? \\$$

Commented by mokys last updated on 29/Mar/24

$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \:\frac{−\mathrm{2}{e}^{−\mathrm{2}{x}} }{\mathrm{1}+{e}^{−\mathrm{2}{x}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{du}}{\mathrm{1}+{u}} \\$$$$\\$$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({ln}\mathrm{2}\right) \\$$$$\\$$$$\:{Aldolaimy}\:{Mohammad} \\$$

Commented by MetaLahor1999 last updated on 29/Mar/24

$${thank}\:{u} \\$$

Commented by mokys last updated on 29/Mar/24

$${welcome} \\$$

Answered by lepuissantcedricjunior last updated on 29/Mar/24

$$\int_{\mathrm{0}} ^{\infty} \frac{{d}\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} }=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} }{\mathrm{1}+\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} }\boldsymbol{{dx}}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{2}\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} }{\mathrm{1}+\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} }\boldsymbol{{dx}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} \right)\right]_{\mathrm{0}} ^{\infty} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{ln}}\sqrt{\mathrm{2}} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} }=\boldsymbol{{ln}}\left(\sqrt{\mathrm{2}}\right) \\$$$$===================== \\$$$$.......{le}\:{puissant}\:\boldsymbol{{D}}{r}....................... \\$$

Answered by mathzup last updated on 31/Mar/24

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