Question Number 132641 by aurpeyz last updated on 15/Feb/21
$$\int\frac{\mathrm{1}}{{e}^{{x}} +\mathrm{9}{e}^{−{x}} }{dx} \\ $$
Answered by MJS_new last updated on 15/Feb/21
$${t}=\mathrm{e}^{{x}} \:\rightarrow\:{dx}=\frac{{dt}}{{t}} \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:\frac{{t}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\:\frac{\mathrm{e}^{{x}} }{\mathrm{3}}\:+{C} \\ $$
Answered by physicstutes last updated on 15/Feb/21
$$\:\mathcal{I}\:=\:\int\frac{\mathrm{1}}{{e}^{{x}} +\mathrm{9}{e}^{−{x}} }\:{dx}\:=\:\int\frac{{e}^{{x}} {dx}}{{e}^{\mathrm{2}{x}} +\:\mathrm{3}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{let}\:{e}^{{x}} \:=\:{u}\:\Rightarrow\:{du}\:=\:{e}^{{x}} {dx} \\ $$$$\Rightarrow\:\mathcal{I}\:=\:\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{3}}\right)+\:{A} \\ $$$$\mathcal{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\mathrm{3}}\right)\:+\:{A}\:,\:{A}\:\in\:\mathbb{R} \\ $$