Question Number 20164 by tammi last updated on 23/Aug/17
$$\int\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} {x}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$
Answered by ajfour last updated on 23/Aug/17
$${let}\:\mathrm{tan}^{−\mathrm{1}} {x}={t} \\ $$$$\Rightarrow\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }={dt} \\ $$$$\Rightarrow\:\:\int\frac{\:\:{e}^{\mathrm{tan}^{−\mathrm{1}} {x}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\int{e}^{{t}} {dt}\:={e}^{{t}} +{C} \\ $$$$\:\:\:\:\:={e}^{\mathrm{tan}^{−\mathrm{1}} {x}} +{C}\:. \\ $$