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Question Number 100191 by bobhans last updated on 25/Jun/20

∫ x^2 e^x dx ?

$$\int\:{x}^{\mathrm{2}} \:{e}^{{x}} \:{dx}\:? \\ $$

Commented by bobhans last updated on 25/Jun/20

D.I method ⇒ x^2 e^x −2xe^x +2e^x +C D I x^2 (+) e^x 2 (−) e^x 2 (+) e^x 0 (−) e^x

$$\mathrm{D}.\mathrm{I}\:\mathrm{method}\:\Rightarrow\:{x}^{\mathrm{2}} {e}^{{x}} −\mathrm{2}{xe}^{{x}} +\mathrm{2}{e}^{{x}} \:+{C} \\ $$$$\:\mathrm{D}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I} \\ $$$${x}^{\mathrm{2}} \:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:\:{e}^{{x}} \: \\ $$$$\mathrm{2}\:\:\:\:\:\left(−\right)\:\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$$$\mathrm{2}\:\:\:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$$$\mathrm{0}\:\:\:\:\:\:\left(−\right)\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$

Answered by Dwaipayan Shikari last updated on 25/Jun/20

=x^2 ∫e^x −∫2xe^x =x^2 e^x −2xe^x +2e^x +Constant

$$={x}^{\mathrm{2}} \int{e}^{{x}} −\int\mathrm{2}{xe}^{{x}} ={x}^{\mathrm{2}} {e}^{{x}} −\mathrm{2}{xe}^{{x}} +\mathrm{2}{e}^{{x}} +\mathrm{Constant} \\ $$$$ \\ $$

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