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How do you find the integral of #x^2 e^(-x) dx#?

1 Answer

Jul 7, 2015

#-x^2 e^-x -2xe^-x -2e^-x # +C

Explanation:

This can be solved, intergrating it by parts,

I= #-x^2 e^-x - int 2x (-e^-x) dx#

= #-x^2 e^-x +2 int xe^-x dx#

=#-x^2 e^-x +2( -xe^-x - int -e^-xdx )#

=#-x^2 e^-x -2xe^-x -2e^-x # +C

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