Integration Questions

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

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

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

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

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

$${set}\:{e}^{{x}} \:=\:\mathrm{tan}\:^{\mathrm{2}} {p}\:\rightarrow{dx}\:=\:\frac{\mathrm{2sec}\:^{\mathrm{2}} {p}\:{dp}}{\mathrm{tan}\:{p}} \\$$$${I}=\:\mathrm{2}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {p}\:{dp}}{\mathrm{tan}\:{p}\left(\mathrm{tan}\:^{\mathrm{2}} {p}+\mathrm{1}\right)} \\$$$${I}\:=\:\mathrm{2}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cot}\:{p}\:{dp}\:=\:\mathrm{2}\left[\:\mathrm{ln}\:\mid\mathrm{sin}\:{p}\mid\right]_{\pi/\mathrm{4}} ^{\pi/\mathrm{2}} \\$$$$=\:\mathrm{2ln}\left(\mathrm{1}\right)−\mathrm{2ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:=\:\mathrm{ln}\left(\mathrm{2}\right)\: \\$$$$\left(\mathrm{J}{S}\:\circledast\right) \\$$

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

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{x}} \:+\mathrm{1}}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{dt}}{\mathrm{t}\left(\mathrm{t}+\mathrm{1}\right)} \\$$$$=\int_{\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt}\:=\left[\mathrm{ln}\mid\frac{\mathrm{t}}{\mathrm{t}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{\infty} \:=−\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{ln}\left(\mathrm{2}\right) \\$$