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Question Number 126989 by arash sharifi last updated on 25/Dec/20

∫e^(√x) dx

$$\int{e}^{\sqrt{{x}}} {dx} \\ $$

Answered by bramlexs22 last updated on 26/Dec/20

let (√x) = u ⇒x = u^2          dx = 2u du   I= ∫ 2u e^u  du ...(by parts )  I =2ue^u −2e^u  + c   I=2(√x) e^(√x)  −2e^(√x)  + c

$${let}\:\sqrt{{x}}\:=\:{u}\:\Rightarrow{x}\:=\:{u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{dx}\:=\:\mathrm{2}{u}\:{du}\: \\ $$$${I}=\:\int\:\mathrm{2}{u}\:{e}^{{u}} \:{du}\:...\left({by}\:{parts}\:\right) \\ $$$${I}\:=\mathrm{2}{ue}^{{u}} −\mathrm{2}{e}^{{u}} \:+\:{c}\: \\ $$$${I}=\mathrm{2}\sqrt{{x}}\:{e}^{\sqrt{{x}}} \:−\mathrm{2}{e}^{\sqrt{{x}}} \:+\:{c}\: \\ $$

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