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Question Number 40127 by maxmathsup by imad last updated on 16/Jul/18
find the value of ∫_0 ^1 ((e^x −1)/(e^x +1))dx
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} \:+\mathrm{1}}{dx} \\ $$
Commented by math khazana by abdo last updated on 18/Jul/18
let I = ∫_0 ^1 ((e^x −1)/(e^x +1))dx I = ∫_0 ^1 (e^x /(e^x +1))dx −∫_0 ^1 (dx/(e^x +1)) but ∫_0 ^1 (e^x /(e^x +1))dx =[ln(e^x +1)]_0 ^1 =ln(1+e)−ln(2) changement e^x =t give ∫_0 ^1 (dx/(e^x +1))dx = ∫_1 ^e (1/(t+1)) (dt/t) =∫_1 ^e ((1/t) −(1/(t+1)))dt =[ln∣ (t/(t+1))∣]_1 ^e = ln((e/(e+1))) +ln(2) ⇒ I =ln(1+e)−ln(2) −ln((e/(e+1)))−ln(2) =2ln(e+1) −2ln(2) −1 .
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} \:+\mathrm{1}}{dx} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{{x}} }{{e}^{{x}} \:+\mathrm{1}}{dx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{{e}^{{x}} \:+\mathrm{1}}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{{x}} }{{e}^{{x}} \:+\mathrm{1}}{dx}\:=\left[{ln}\left({e}^{{x}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:={ln}\left(\mathrm{1}+{e}\right)−{ln}\left(\mathrm{2}\right) \\ $$$${changement}\:{e}^{{x}} ={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{{e}^{{x}} \:+\mathrm{1}}{dx}\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\frac{{dt}}{{t}}\:=\int_{\mathrm{1}} ^{{e}} \left(\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\left[{ln}\mid\:\frac{{t}}{{t}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{{e}} =\:{ln}\left(\frac{{e}}{{e}+\mathrm{1}}\right)\:+{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${I}\:={ln}\left(\mathrm{1}+{e}\right)−{ln}\left(\mathrm{2}\right)\:−{ln}\left(\frac{{e}}{{e}+\mathrm{1}}\right)−{ln}\left(\mathrm{2}\right) \\ $$$$=\mathrm{2}{ln}\left({e}+\mathrm{1}\right)\:−\mathrm{2}{ln}\left(\mathrm{2}\right)\:−\mathrm{1}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18
∫_0 ^1 ((e^x −1)/(e^x +1))dx t=e^x +1 dt=e^x dx (dt/(t−1))=dx ∫_2 ^(e+1) ((t−2)/t).(dt/(t−1)) ∫_2 ^(e+1) ((2(t−1)−t)/(t(t−1)))dt ∫_2 ^(e+1) ((2dt)/t)−∫_2 ^(e+1) (dt/(t−1)) 2lnt−ln(t−1)∣_2 ^(e+1) =∣ln(t^2 /(t−1))∣_2 ^(e+1) =ln(((e+1)^2 )/(e+1−1)) −ln(2^2 /(2−1)) =2ln(e+1)−lne−2ln2 =
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$${t}={e}^{{x}} +\mathrm{1}\:\:{dt}={e}^{{x}} {dx} \\ $$$$\frac{{dt}}{{t}−\mathrm{1}}={dx} \\ $$$$\int_{\mathrm{2}} ^{{e}+\mathrm{1}} \:\frac{{t}−\mathrm{2}}{{t}}.\frac{{dt}}{{t}−\mathrm{1}} \\ $$$$\int_{\mathrm{2}} ^{{e}+\mathrm{1}} \:\:\frac{\mathrm{2}\left({t}−\mathrm{1}\right)−{t}}{{t}\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\int_{\mathrm{2}} ^{{e}+\mathrm{1}} \:\frac{\mathrm{2}{dt}}{{t}}−\int_{\mathrm{2}} ^{{e}+\mathrm{1}} \:\frac{{dt}}{{t}−\mathrm{1}} \\ $$$$\:\:\mathrm{2}{lnt}−{ln}\left({t}−\mathrm{1}\right)\mid_{\mathrm{2}} ^{{e}+\mathrm{1}} \\ $$$$=\mid{ln}\frac{{t}^{\mathrm{2}} }{{t}−\mathrm{1}}\mid_{\mathrm{2}} ^{{e}+\mathrm{1}} \\ $$$$={ln}\frac{\left({e}+\mathrm{1}\right)^{\mathrm{2}} }{{e}+\mathrm{1}−\mathrm{1}}\:−{ln}\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}−\mathrm{1}} \\ $$$$=\mathrm{2}{ln}\left({e}+\mathrm{1}\right)−{lne}−\mathrm{2}{ln}\mathrm{2} \\ $$$$ \\ $$$$= \\ $$

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