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