Question Number 151753 by Integrals last updated on 22/Aug/21
Commented by puissant last updated on 22/Aug/21
![K=∫(√x)e^(√x) dx u=(√x)→ u^2 =x → dx=2udu K=2∫u^2 e^u du { ((i=u^2 )),((j′=e^u )) :} ⇒ { ((i′=2u)),((j=e^u )) :} K=2[u^2 e^u ]−4∫ue^u du { ((i=u)),((j′=e^u )) :} ⇒ { ((i′=1)),((j=e^u )) :} ⇒ K=2u^2 e^u −4ue^u +4e^u +C K=2xe^(√x) −4(√x)e^(√x) +4e^(√x) +C..](Q151774.png)
$${K}=\int\sqrt{{x}}{e}^{\sqrt{{x}}} {dx} \\ $$$${u}=\sqrt{{x}}\rightarrow\:{u}^{\mathrm{2}} ={x}\:\rightarrow\:{dx}=\mathrm{2}{udu} \\ $$$${K}=\mathrm{2}\int{u}^{\mathrm{2}} {e}^{{u}} {du} \\ $$$$\begin{cases}{{i}={u}^{\mathrm{2}} }\\{{j}'={e}^{{u}} }\end{cases}\:\:\Rightarrow\:\:\begin{cases}{{i}'=\mathrm{2}{u}}\\{{j}={e}^{{u}} }\end{cases} \\ $$$${K}=\mathrm{2}\left[{u}^{\mathrm{2}} {e}^{{u}} \right]−\mathrm{4}\int{ue}^{{u}} {du} \\ $$$$\begin{cases}{{i}={u}}\\{{j}'={e}^{{u}} }\end{cases}\:\:\Rightarrow\:\begin{cases}{{i}'=\mathrm{1}}\\{{j}={e}^{{u}} }\end{cases} \\ $$$$\Rightarrow\:{K}=\mathrm{2}{u}^{\mathrm{2}} {e}^{{u}} −\mathrm{4}{ue}^{{u}} +\mathrm{4}{e}^{{u}} +{C} \\ $$$${K}=\mathrm{2}{xe}^{\sqrt{{x}}} −\mathrm{4}\sqrt{{x}}{e}^{\sqrt{{x}}} +\mathrm{4}{e}^{\sqrt{{x}}} +{C}.. \\ $$