How do you integrate #int x^2 e^(-x) dx # using integration by parts?

1 Answer
Lovecraft
Jan 9, 2016

#I = -e^(-x)(x^2 + 2x + 2) + c#

Explanation:

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

Say #u = x^2# so #du = 2x# and #dv = e^(-x)# so #v = -e^(-x)#

#I = -x^2e^(-x) + 2intxe^(-x)dx#

Say #u = x# so #du = 1# and #dv = e^(-x)# so #v = -e^(-x)#

#I = -x^2e^(-x) + 2(-xe^(-x) + inte^(-x)dx)#
#I = -x^2e^(-x) -2xe^(-x) + 2inte^(-x)dx#
#I = -x^2e^(-x)-2xe^(-x)-2e^(-x)+c#
#I = -e^(-x)(x^2 + 2x + 2) + c#

Related questions
See all questions in Integration by Parts
Impact of this question
1482 views around the world
You can reuse this answer
Creative Commons License