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Question-168952




Question Number 168952 by Sotoberry last updated on 22/Apr/22
Answered by alephzero last updated on 22/Apr/22
∫((2+e^x )/e^x ) dx  u = e^x   du = e^x  dx  dx = (du/e^x ) = (du/u)  ⇒ ∫((2+e^x )/e^x )dx = ∫((2+u)/u) (du/u) =  = ∫((2+u)/u^2 ) du = ∫(2/u^2 )du+∫(1/u)du =  = 2∫u^(−2) du+ln∣u∣ =  = 2(−1u^(−1) )+ln∣u∣+c  = −(2/u)+ln∣u∣+c = −(2/e^x )+ln∣e^x ∣+c  ∀x ∈ R e^x  > 0  ⇒ −(2/e^x )+ln∣e^x ∣+c = −(2/e^x )+x+c  ⇒ ∫(((2+e^x )dx)/e^x ) = x−(2/e^x )+c
$$\int\frac{\mathrm{2}+{e}^{{x}} }{{e}^{{x}} }\:{dx} \\ $$$${u}\:=\:{e}^{{x}} \\ $$$${du}\:=\:{e}^{{x}} \:{dx} \\ $$$${dx}\:=\:\frac{{du}}{{e}^{{x}} }\:=\:\frac{{du}}{{u}} \\ $$$$\Rightarrow\:\int\frac{\mathrm{2}+{e}^{{x}} }{{e}^{{x}} }{dx}\:=\:\int\frac{\mathrm{2}+{u}}{{u}}\:\frac{{du}}{{u}}\:= \\ $$$$=\:\int\frac{\mathrm{2}+{u}}{{u}^{\mathrm{2}} }\:{du}\:=\:\int\frac{\mathrm{2}}{{u}^{\mathrm{2}} }{du}+\int\frac{\mathrm{1}}{{u}}{du}\:= \\ $$$$=\:\mathrm{2}\int{u}^{−\mathrm{2}} {du}+\mathrm{ln}\mid{u}\mid\:= \\ $$$$=\:\mathrm{2}\left(−\mathrm{1}{u}^{−\mathrm{1}} \right)+\mathrm{ln}\mid{u}\mid+{c} \\ $$$$=\:−\frac{\mathrm{2}}{{u}}+\mathrm{ln}\mid{u}\mid+{c}\:=\:−\frac{\mathrm{2}}{{e}^{{x}} }+\mathrm{ln}\mid{e}^{{x}} \mid+{c} \\ $$$$\forall{x}\:\in\:\mathbb{R}\:{e}^{{x}} \:>\:\mathrm{0} \\ $$$$\Rightarrow\:−\frac{\mathrm{2}}{{e}^{{x}} }+\mathrm{ln}\mid{e}^{{x}} \mid+{c}\:=\:−\frac{\mathrm{2}}{{e}^{{x}} }+{x}+{c} \\ $$$$\Rightarrow\:\int\frac{\left(\mathrm{2}+{e}^{{x}} \right){dx}}{{e}^{{x}} }\:=\:{x}−\frac{\mathrm{2}}{{e}^{{x}} }+{c} \\ $$

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