How do you find the integral of $x^{2} e^{- x} d x$ $x^2 e^(-x) dx$ ?

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Answer 1 · 2015-07-07T04:05:28

$- x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x}$ $-x^2 e^-x -2xe^-x -2e^-x$ +C

This can be solved, intergrating it by parts,

I= $- x^{2} e^{- x} - \int 2 x (- e^{- x}) d x$ $-x^2 e^-x - int 2x (-e^-x) dx$

= $- x^{2} e^{- x} + 2 \int x e^{- x} d x$ $-x^2 e^-x +2 int xe^-x dx$

= $- x^{2} e^{- x} + 2 (- x e^{- x} - \int - e^{- x} d x)$ $-x^2 e^-x +2( -xe^-x - int -e^-xdx )$

= $- x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x}$ $-x^2 e^-x -2xe^-x -2e^-x$ +C

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