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Question Number 101835 by bobhans last updated on 05/Jul/20

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{e^x+1}dx$$

Answered by Dwaipayan Shikari last updated on 05/Jul/20

$${\int }_0^{\mathrm{\infty }}\frac{e^x}{e^x(e^x+1)}dx={\int }_0^{\mathrm{\infty }}\frac{dt}{(t-1)t}={\int }_2^{\mathrm{\infty }}(\frac{1}{t-1}-\frac{1}{t})dt={\left[log\left(\frac{t-1}{t}\right)\right]}_2^{\mathrm{\infty }}$$$$=log2\{suppose{e}^x+1=t$$

Answered by john santu last updated on 05/Jul/20

$$set{e}^x=\tan {}^{2}p\to dx=\frac{2\sec {}^{2}pdp}{\tan p}$$$$I=2\underset{\pi /4}{\stackrel{\pi /2}{\int }}\frac{\sec {}^{2}pdp}{\tan p(\tan {}^{2}p+1)}$$$$I=2\underset{\pi /4}{\stackrel{\pi /2}{\int }}\cot pdp=2{[\ln \mid \sin p\mid ]}_{\pi /4}^{\pi /2}$$$$=2\ln \left(1\right)-2\ln \left(\frac{\sqrt{2}}{2}\right)=\ln \left(2\right)$$$$(\mathrm{J}S⊛)$$

Answered by mathmax by abdo last updated on 05/Jul/20

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

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