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Question-31941




Question Number 31941 by mondodotto@gmail.com last updated on 17/Mar/18
Commented by abdo imad last updated on 20/Mar/18
the ch. e^x =t give x=ln(t) and  ∫(dx/(1+e^x )) =  ∫   (1/(1+t)) (dt/t) = ∫ ((1/t) −(1/(t+1)))dt  = ln∣t∣−ln∣t+1∣ +λ   = x −ln(e^x  +1) +λ  .
$${the}\:{ch}.\:{e}^{{x}} ={t}\:{give}\:{x}={ln}\left({t}\right)\:{and} \\ $$$$\int\frac{{dx}}{\mathrm{1}+{e}^{{x}} }\:=\:\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\frac{{dt}}{{t}}\:=\:\int\:\left(\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\:{ln}\mid{t}\mid−{ln}\mid{t}+\mathrm{1}\mid\:+\lambda\: \\ $$$$=\:{x}\:−{ln}\left({e}^{{x}} \:+\mathrm{1}\right)\:+\lambda\:\:. \\ $$
Answered by Joel578 last updated on 17/Mar/18
I = ∫ (1/(1 + e^x )) dx = ∫ ((1 + e^x  − e^x )/(1 + e^x )) dx     = ∫ 1 − (e^x /(1 + e^x )) dx  u = e^x   →  du = e^x  dx  I = ∫ dx − ∫ (u/(1 + u)) (du/u)     = x − ln ∣1 + u∣ + C     = x − ln ∣1 + e^x ∣ + C
$${I}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{e}^{{x}} }\:{dx}\:=\:\int\:\frac{\mathrm{1}\:+\:{e}^{{x}} \:−\:{e}^{{x}} }{\mathrm{1}\:+\:{e}^{{x}} }\:{dx} \\ $$$$\:\:\:=\:\int\:\mathrm{1}\:−\:\frac{{e}^{{x}} }{\mathrm{1}\:+\:{e}^{{x}} }\:{dx} \\ $$$${u}\:=\:{e}^{{x}} \:\:\rightarrow\:\:{du}\:=\:{e}^{{x}} \:{dx} \\ $$$${I}\:=\:\int\:{dx}\:−\:\int\:\frac{{u}}{\mathrm{1}\:+\:{u}}\:\frac{{du}}{{u}} \\ $$$$\:\:\:=\:{x}\:−\:\mathrm{ln}\:\mid\mathrm{1}\:+\:{u}\mid\:+\:{C} \\ $$$$\:\:\:=\:{x}\:−\:\mathrm{ln}\:\mid\mathrm{1}\:+\:{e}^{{x}} \mid\:+\:{C} \\ $$
Commented by mondodotto@gmail.com last updated on 19/Mar/18
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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