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

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