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e-tan-1-x-1-x-2-




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

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