[tex]\displaystyle\mathrm{I = \int_{0}^1(x^2+2x+1)e^x\,dx = \int_{0}^1(x+1)^2e^x\, dx} = \\ \\ =\mathrm{\int_{0}^1(x+1)^2(e^x)'\, dx = (x+1)^2e^x\Big|_{0}^1 - \int_{0}^1 \Big[(x+1)^2\Big]'e^x\, dx =}\\ \\ = \mathrm{4e-1 - \int_{0}^1 2(x+1)e^x\, dx = 4e-1-2\int_{0}^1(x+1)(e^x)'\, dx =}\\ \\ \mathrm{= 4e-1-\Big[2(x+1)e^x\Big]\Big|_{0}^1+2\int_{0}^1(x+1)'e^x\, dx =} \\ \\ \mathrm{= 4e-1-4e+2+2\int_{0}^1 e^x\, dx = 1+2(e^x)\Big|_{0}^1 = }\\ \\ \mathrm{=1+2e-2 = \boxed{2e-1}}[/tex]
