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Question Number 51420 by Tawa1 last updated on 26/Dec/18

∫  (e^(3x) /(1 + e^x )) dx

$$\int\:\:\frac{\mathrm{e}^{\mathrm{3x}} }{\mathrm{1}\:+\:\mathrm{e}^{\mathrm{x}} }\:\mathrm{dx} \\ $$

Answered by ajfour last updated on 26/Dec/18

let  1+e^x  = t   , ⇒  e^x dx = dt  and  e^(3x)  dx = e^(2x) (e^x dx) = (t−1)^2 dt  I = ∫(((t−1)^2 dt)/t) = ∫(t−2+(1/t))dt  ⇒   I = (((1+e^x )^2 )/2)−2e^x +ln (1+e^x )+c .

$${let}\:\:\mathrm{1}+{e}^{{x}} \:=\:{t}\:\:\:,\:\Rightarrow\:\:{e}^{{x}} {dx}\:=\:{dt} \\ $$$${and}\:\:{e}^{\mathrm{3}{x}} \:{dx}\:=\:{e}^{\mathrm{2}{x}} \left({e}^{{x}} {dx}\right)\:=\:\left({t}−\mathrm{1}\right)^{\mathrm{2}} {dt} \\ $$$${I}\:=\:\int\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} {dt}}{{t}}\:=\:\int\left({t}−\mathrm{2}+\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$$\Rightarrow\:\:\:{I}\:=\:\frac{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{e}^{{x}} +\mathrm{ln}\:\left(\mathrm{1}+{e}^{{x}} \right)+{c}\:. \\ $$

Commented by Tawa1 last updated on 26/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Smail last updated on 26/Dec/18

let  t=e^x ⇒e^x dx=dt  I=∫(e^(3x) /(1+e^x ))dx=∫(t^2 /(1+t))dt=∫((t(t+1)−(t+1)+1)/(1+t))dt  =∫(t−1+(1/(1+t)))dt  =(t^2 /2)−t+ln∣t+1∣+C  I=(e^(2x) /2)−e^x +ln(e^x +1)+C

$${let}\:\:{t}={e}^{{x}} \Rightarrow{e}^{{x}} {dx}={dt} \\ $$$${I}=\int\frac{{e}^{\mathrm{3}{x}} }{\mathrm{1}+{e}^{{x}} }{dx}=\int\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}}{dt}=\int\frac{{t}\left({t}+\mathrm{1}\right)−\left({t}+\mathrm{1}\right)+\mathrm{1}}{\mathrm{1}+{t}}{dt} \\ $$$$=\int\left({t}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−{t}+{ln}\mid{t}+\mathrm{1}\mid+{C} \\ $$$${I}=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{2}}−{e}^{{x}} +{ln}\left({e}^{{x}} +\mathrm{1}\right)+{C} \\ $$

Commented by Tawa1 last updated on 26/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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