Question Number 100191 by bobhans last updated on 25/Jun/20
$$\int\:{x}^{\mathrm{2}} \:{e}^{{x}} \:{dx}\:? \\ $$
Commented by bobhans last updated on 25/Jun/20
$$\mathrm{D}.\mathrm{I}\:\mathrm{method}\:\Rightarrow\:{x}^{\mathrm{2}} {e}^{{x}} −\mathrm{2}{xe}^{{x}} +\mathrm{2}{e}^{{x}} \:+{C} \\ $$$$\:\mathrm{D}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I} \\ $$$${x}^{\mathrm{2}} \:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:\:{e}^{{x}} \: \\ $$$$\mathrm{2}\:\:\:\:\:\left(−\right)\:\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$$$\mathrm{2}\:\:\:\:\:\left(+\right)\:\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$$$\mathrm{0}\:\:\:\:\:\:\left(−\right)\:\:\:\:\:\:\:\:{e}^{{x}} \\ $$
Answered by Dwaipayan Shikari last updated on 25/Jun/20
$$={x}^{\mathrm{2}} \int{e}^{{x}} −\int\mathrm{2}{xe}^{{x}} ={x}^{\mathrm{2}} {e}^{{x}} −\mathrm{2}{xe}^{{x}} +\mathrm{2}{e}^{{x}} +\mathrm{Constant} \\ $$$$ \\ $$