e-x-dx-

Question Number 154104 by ZiYangLee last updated on 14/Sep/21

$$\:\:\int\:{e}^{\sqrt{{x}}} \:{dx}\:=? \\ $$

Answered by peter frank last updated on 14/Sep/21

$${u}=\sqrt{{x}} \\ $$$${u}^{\mathrm{2}} ={x} \\ $$$$\mathrm{2}{udu}={dx} \\ $$$$\int{e}^{{u}} \mathrm{2}{udu} \\ $$$$\mathrm{2}\int{ue}^{{u}} \\ $$$${use}\:{by}\:−{part} \\ $$

Commented by ZiYangLee last updated on 14/Sep/21

$${i}\:{see}…\:{thanks} \\ $$

Answered by puissant last updated on 14/Sep/21

$${u}=\sqrt{{x}}\:\rightarrow\:{u}^{\mathrm{2}} ={x}\:\rightarrow\:\mathrm{2}{udu}={dx} \\ $$$${Q}=\int{e}^{\sqrt{{x}}} {dx}\:=\:\mathrm{2}\int{ue}^{{u}} {du} \\ $$$$\begin{cases}{{i}={u}}\\{{j}'={e}^{{u}} }\end{cases}\Rightarrow\:\begin{cases}{{i}'=\mathrm{1}}\\{{j}={e}^{{u}} }\end{cases} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\left[{ue}^{{u}} \right]−\mathrm{2}\int{e}^{{u}} {du} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}{ue}^{{u}} −\mathrm{2}{e}^{{u}} +{C} \\ $$$$ \\ $$$$\therefore\because\:\:{Q}\:=\:\mathrm{2}\sqrt{{x}}{e}^{\sqrt{{x}}} −\mathrm{2}{e}^{\sqrt{{x}}} +{C}.. \\ $$

Commented by ZiYangLee last updated on 14/Sep/21

$${appreciated}.. \\ $$

Answered by peter frank last updated on 14/Sep/21

$$\mathrm{e}^{\mathrm{x}} =\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\int\left(\mathrm{1}+\left(\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{\mathrm{x}}{\mathrm{2}!}+\frac{\left(\mathrm{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:}{\mathrm{3}!}+…\right)\mathrm{dx} \\ $$$$\mathrm{x}+\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}}+\frac{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}} }{\mathrm{6}×\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}+…+\mathrm{A} \\ $$$$ \\ $$

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