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Question Number 172011 by Mikenice last updated on 23/Jun/22
find integrate:  ∫x^2 e^x dx
$${find}\:{integrate}: \\ $$$$\int{x}^{\mathrm{2}} {e}^{{x}} {dx} \\ $$
Answered by puissant last updated on 23/Jun/22
P = ∫x^2 e^x dx        { ((u′=e^x )),((v=x^2 )) :}          ⇒       { ((u=e^x )),((v′= 2x)) :}   P= x^2 e^x  − 2∫xe^x dx   P=x^2 e^x  − 2xe^x +2e^x +C   P = (x^2 −2x+2)e^x +C
$${P}\:=\:\int{x}^{\mathrm{2}} {e}^{{x}} {dx} \\ $$$$\:\:\:\:\:\begin{cases}{{u}'={e}^{{x}} }\\{{v}={x}^{\mathrm{2}} }\end{cases}\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\begin{cases}{{u}={e}^{{x}} }\\{{v}'=\:\mathrm{2}{x}}\end{cases} \\ $$$$\:{P}=\:{x}^{\mathrm{2}} {e}^{{x}} \:−\:\mathrm{2}\int{xe}^{{x}} {dx} \\ $$$$\:{P}={x}^{\mathrm{2}} {e}^{{x}} \:−\:\mathrm{2}{xe}^{{x}} +\mathrm{2}{e}^{{x}} +{C} \\ $$$$\:{P}\:=\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right){e}^{{x}} +{C} \\ $$

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