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Question Number 156600 by ZiYangLee last updated on 13/Oct/21

∫ (e^x /(e^(2x) +4)) dx =?

$$\int\:\frac{{e}^{{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{4}}\:{dx}\:=?\: \\ $$

Answered by gsk2684 last updated on 13/Oct/21

put e^x =t e^x dx=dt ∫(1/(t^2 +2^2 ))dt =(1/2)tan^(−1) ((t/2))+c =(1/2)tan^(−1) ((e^x /2))+c

$${put}\:{e}^{{x}} ={t} \\ $$$${e}^{{x}} \:{dx}={dt} \\ $$$$\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}}{\mathrm{2}}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\mathrm{2}}\right)+{c}\: \\ $$

Answered by physicstutes last updated on 13/Oct/21

Another cool approach. ∫(e^x /(e^(2x) +4))dx = ∫(1/(e^(2x) +4)) e^x dx let u = e^x ⇒ du = e^x dx ⇒ ∫(du/(u^2 +4))du = (1/2)tan^(−1) ((u/2))+k = (1/2) tan^(−1) ((e^x /2))+k

$$\mathrm{Another}\:\mathrm{cool}\:\mathrm{approach}. \\ $$$$\int\frac{{e}^{{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{4}}{dx}\:\:=\:\int\frac{\mathrm{1}}{{e}^{\mathrm{2}{x}} +\mathrm{4}}\:{e}^{{x}} {dx} \\ $$$$\mathrm{let}\:{u}\:=\:{e}^{{x}} \:\Rightarrow\:{du}\:=\:{e}^{{x}} {dx} \\ $$$$\Rightarrow\:\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{4}}{du}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{2}}\right)+{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\mathrm{2}}\right)+{k} \\ $$

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