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Question Number 35920 by rahul 19 last updated on 25/May/18

∫ ((e^(2x) +1)/(2e^x −1)) dx = ?

$$\int\:\frac{{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{2}{e}^{{x}} −\mathrm{1}}\:{dx}\:=\:? \\ $$

Commented by prof Abdo imad last updated on 26/May/18

changement e^x =t give I = ∫ ((t^2 +1)/(2t−1)) (dt/t) = ∫ ((t^2 +1)/(2t^2 −t))dt =(1/2)∫ ((2t^2 +2 −t +t)/(2t^2 −t))dt = (t/2) + (1/2)∫ ((2+t)/(2t^2 −t))dt let decompose F(t) = ((2+t)/(t(2t−1))) =(a/t) +((bt +c)/(2t−1)) ⇔ 2at −a +bt^2 +ct =2+t ⇔ bt^2 +(2a+c)t −a =2+t ⇒b=0 , a =−2 , 2a+c=1 ⇒c=1−2(−2)=5 F(t)= ((−2)/t) +(5/(2t−1)) ∫ ((2+t)/(2t^2 −t))dt = −2ln∣t∣ +(5/2)ln∣t−(1/2)∣ +c ⇒ I = (t/2) −ln∣t∣ +(5/4)ln∣t−(1/2)∣ +c ⇒ I = (e^x /2) −x +(5/4)ln∣e^x −(1/2)∣ +c .

$${changement}\:{e}^{{x}} \:={t}\:{give}\: \\ $$$${I}\:=\:\int\:\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}−\mathrm{1}}\:\frac{{dt}}{{t}}\:=\:\int\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} \:−{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}\:−{t}\:+{t}}{\mathrm{2}{t}^{\mathrm{2}} \:−{t}}{dt} \\ $$$$=\:\frac{{t}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{2}+{t}}{\mathrm{2}{t}^{\mathrm{2}} \:−{t}}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\:\frac{\mathrm{2}+{t}}{{t}\left(\mathrm{2}{t}−\mathrm{1}\right)}\:=\frac{{a}}{{t}}\:+\frac{{bt}\:+{c}}{\mathrm{2}{t}−\mathrm{1}} \\ $$$$\Leftrightarrow\:\mathrm{2}{at}\:−{a}\:+{bt}^{\mathrm{2}} \:+{ct}\:=\mathrm{2}+{t}\:\Leftrightarrow\:{bt}^{\mathrm{2}} \:+\left(\mathrm{2}{a}+{c}\right){t}\:−{a} \\ $$$$=\mathrm{2}+{t}\:\Rightarrow{b}=\mathrm{0}\:,\:{a}\:=−\mathrm{2}\:,\:\mathrm{2}{a}+{c}=\mathrm{1}\:\Rightarrow{c}=\mathrm{1}−\mathrm{2}\left(−\mathrm{2}\right)=\mathrm{5} \\ $$$${F}\left({t}\right)=\:\:\frac{−\mathrm{2}}{{t}}\:\:+\frac{\mathrm{5}}{\mathrm{2}{t}−\mathrm{1}} \\ $$$$\int\:\:\frac{\mathrm{2}+{t}}{\mathrm{2}{t}^{\mathrm{2}} \:−{t}}{dt}\:=\:−\mathrm{2}{ln}\mid{t}\mid\:\:+\frac{\mathrm{5}}{\mathrm{2}}{ln}\mid{t}−\frac{\mathrm{1}}{\mathrm{2}}\mid\:+{c}\:\Rightarrow \\ $$$${I}\:=\:\frac{{t}}{\mathrm{2}}\:−{ln}\mid{t}\mid\:+\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{t}−\frac{\mathrm{1}}{\mathrm{2}}\mid\:+{c}\:\:\Rightarrow \\ $$$${I}\:=\:\frac{{e}^{{x}} }{\mathrm{2}}\:−{x}\:+\frac{\mathrm{5}}{\mathrm{4}}{ln}\mid{e}^{{x}} \:−\frac{\mathrm{1}}{\mathrm{2}}\mid\:+{c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18

t=2e^x −1 e^x =((t+1)/2) so e^x dx=(1/2)dt dx=(1/2)×(2/(t+1))×dt ∫(((((t+1)/2))^2 +1)/t)×(dt/(t+1)) ∫((t^2 +2t+1+4)/(4t(t+1)))dt (1/4)∫((t^2 +2t+5)/(t(t+1)))dt deviding t^2 +2t+5 by t^2 +t (1/4)∫{1+((t+5)/(t^2 +t))}dt (1/4)∫tdt+(1/8)∫((2t+1+9)/(t^2 +t))dt (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +t))dt+(9/8)∫(dt/(t(t+1))) (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +1))dt+(9/8)∫((t+1−t)/(t(t+1)))dt (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +t))dt+(9/8)∫(dt/t)−(9/8)∫(dt/(t+1)) (1/4)t+(1/8)ln(t^2 +t)+(9/8)lnt−(9/8)ln(t+1) (1/4)(2e^x −1)+(1/8)ln{(2e^x −1)^2 +(2e^x −1}− (9/8)ln(2e^x −1+1) (1/4)(2e^x −1)+(1/8)ln(4e^(2x) −4e^x +1+2e^x −1)− (9/8){ln2+x} (1/4)(2^ e^x −1)+(1/8)ln(4e^(2x) −2e^x )+((9x)/8)+constant (1/4)(2e^x −1)+(1/8){ln2e^x (2e^x −1)}+((9x)/8)+C =(1/4)(2e^x −1)+(1/8){ln2+x+ln(2e^x −1)}+((9x)/8)+C =(1/4)(2e^x −1)+((ln2)/8)+(x/8)+(1/8)ln(2e^x −1)+((9x)/8)+C =(1/4)(2e^x −1)+((10x)/8)+(1/8)ln(2e^x −1)+C′ ans

$${t}=\mathrm{2}{e}^{{x}} −\mathrm{1}\: \\ $$$${e}^{{x}} =\frac{{t}+\mathrm{1}}{\mathrm{2}}\:\:{so}\:{e}^{{x}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}{dt} \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{{t}+\mathrm{1}}×{dt} \\ $$$$\int\frac{\left(\frac{{t}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}}{{t}}×\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$\int\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}+\mathrm{4}}{\mathrm{4}{t}\left({t}+\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{5}}{{t}\left({t}+\mathrm{1}\right)}{dt} \\ $$$${deviding}\:{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{5}\:\:{by}\:{t}^{\mathrm{2}} +{t} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\left\{\mathrm{1}+\frac{{t}+\mathrm{5}}{{t}^{\mathrm{2}} +{t}}\right\}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{tdt}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{2}{t}+\mathrm{1}+\mathrm{9}}{{t}^{\mathrm{2}} +{t}}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{dt}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}}{dt}+\frac{\mathrm{9}}{\mathrm{8}}\int\frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{dt}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\frac{\mathrm{9}}{\mathrm{8}}\int\frac{{t}+\mathrm{1}−{t}}{{t}\left({t}+\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int{dt}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}}{dt}+\frac{\mathrm{9}}{\mathrm{8}}\int\frac{{dt}}{{t}}−\frac{\mathrm{9}}{\mathrm{8}}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{t}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({t}^{\mathrm{2}} +{t}\right)+\frac{\mathrm{9}}{\mathrm{8}}{lnt}−\frac{\mathrm{9}}{\mathrm{8}}{ln}\left({t}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left\{\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right\}−\right. \\ $$$$\frac{\mathrm{9}}{\mathrm{8}}{ln}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{4}{e}^{\mathrm{2}{x}} −\mathrm{4}{e}^{{x}} +\mathrm{1}+\mathrm{2}{e}^{{x}} −\mathrm{1}\right)− \\ $$$$\frac{\mathrm{9}}{\mathrm{8}}\left\{{ln}\mathrm{2}+{x}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}^{} {e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{4}{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} \right)+\frac{\mathrm{9}{x}}{\mathrm{8}}+{constant} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}\left\{{ln}\mathrm{2}{e}^{{x}} \left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)\right\}+\frac{\mathrm{9}{x}}{\mathrm{8}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{8}}\left\{{ln}\mathrm{2}+{x}+{ln}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)\right\}+\frac{\mathrm{9}{x}}{\mathrm{8}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{{ln}\mathrm{2}}{\mathrm{8}}+\frac{{x}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{9}{x}}{\mathrm{8}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+\frac{\mathrm{10}{x}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)+{C}'\:\:{ans} \\ $$

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