Question Number 170610 by Sotoberry last updated on 27/May/22
Answered by aleks041103 last updated on 27/May/22
$${IBP} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$${u}={x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$${v}={e}^{{x}} \\ $$$${du}=\left(\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$$${dv}={e}^{{x}} {dx} \\ $$$$\Rightarrow\int\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){e}^{{x}} {dx}=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){e}^{{x}} −\int\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{{x}} {dx} \\ $$$${again}\:{IBP} \\ $$$${u}=\mathrm{2}{x}+\mathrm{1} \\ $$$${v}={e}^{{x}} \\ $$$${du}=\mathrm{2}{dx} \\ $$$${dv}={e}^{{x}} {dx} \\ $$$$\Rightarrow\int\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{{x}} {dx}=\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{{x}} −\mathrm{2}\int{e}^{{x}} {dx}= \\ $$$$=\left(\mathrm{2}{x}−\mathrm{1}\right){e}^{{x}} +{C} \\ $$$$\Rightarrow\int\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right){e}^{{x}} {dx}=\left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right){e}^{{x}} +{C} \\ $$