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=(√x) → u^2 =x → 2udu=dx Q=∫e^(√x) dx = 2∫ue^u du { ((i=u)),((j′=e^u )) :}⇒ { ((i′=1)),((j=e^u )) :} ⇒ Q=2[ue^u ]−2∫e^u du ⇒ Q=2ue^u −2e^u +C ∴∵ Q = 2(√x)e^(√x) −2e^(√x) +C..](https://www.tinkutara.com/question/Q154108.png)
$${u}=\sqrt{{x}}\:\rightarrow\:{u}^{\mathrm{2}} ={x}\:\rightarrow\:\mathrm{2}{udu}={dx} \\ $$$${Q}=\int{e}^{\sqrt{{x}}} {dx}\:=\:\mathrm{2}\int{ue}^{{u}} {du} \\ $$$$\begin{cases}{{i}={u}}\\{{j}'={e}^{{u}} }\end{cases}\Rightarrow\:\begin{cases}{{i}'=\mathrm{1}}\\{{j}={e}^{{u}} }\end{cases} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}\left[{ue}^{{u}} \right]−\mathrm{2}\int{e}^{{u}} {du} \\ $$$$\Rightarrow\:{Q}=\mathrm{2}{ue}^{{u}} −\mathrm{2}{e}^{{u}} +{C} \\ $$$$ \\ $$$$\therefore\because\:\:{Q}\:=\:\mathrm{2}\sqrt{{x}}{e}^{\sqrt{{x}}} −\mathrm{2}{e}^{\sqrt{{x}}} +{C}.. \\ $$
Commented by ZiYangLee last updated on 14/Sep/21
$${appreciated}.. \\ $$
Answered by peter frank last updated on 14/Sep/21
$$\mathrm{e}^{\mathrm{x}} =\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$\int\left(\mathrm{1}+\left(\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{\mathrm{x}}{\mathrm{2}!}+\frac{\left(\mathrm{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:}{\mathrm{3}!}+…\right)\mathrm{dx} \\ $$$$\mathrm{x}+\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} }{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}}+\frac{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}} }{\mathrm{6}×\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}+…+\mathrm{A} \\ $$$$ \\ $$