# Evaluate the integral: int_0^1 (x^2 +1)e^(-x)dx

Evaluate the integral:
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx=$
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Ali Harper
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx={\int }_{0}^{1}{x}^{2}{e}^{-x}dx+{\int }_{0}^{1}{e}^{-x}dx$
${\int }_{0}^{1}{e}^{-x}dx=\left(-{e}^{-x}{\right)}_{0}^{1}=-{e}^{-1}+1$
${\int }_{0}^{1}{x}^{2}{e}^{-x}dx=\left[u={x}^{2}\to {u}^{\prime }=2x,{v}^{\prime }={e}^{-x}\to v=-{e}^{-x}\right]=uv-\int uv=-{x}^{2}{e}^{-x}+\int 2x{e}^{-x}dx$
$\int 2x{e}^{-x}dx=\left[u=2x\to u=2,{v}^{\prime }={e}^{-x}\to v=-{e}^{-x}\right]=-2x{e}^{-x}+\int 2{e}^{-x}dx$
$\to {\int }_{0}^{1}{x}^{2}{e}^{-x}dx=\left(-{x}^{2}{e}^{-x}{\right)}_{0}^{1}-\left(2x{e}^{-x}{\right)}_{0}^{1}+{\int }_{0}^{1}2{e}^{-x}dx=-{e}^{1}-2{e}^{-1}+2\left(-{e}^{-1}+1\right)=-5{e}^{-1}+1$
${\int }_{0}^{1}\left({x}^{2}+1\right){e}^{-x}dx={\int }_{0}^{1}{x}^{2}{e}^{-x}dx+{\int }_{0}^{1}{e}^{-x}dx=-5{e}^{-1}+1-{e}^{-1}+1=-6{e}^{-1}+2$
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