How do you integrate #int x^2e^(x-1)dx# using integration by parts?

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Answer 1 · 2018-06-30T12:35:43

The answer is #=e^(x-1)(x^2-2x+2)+C#

The integral is

#I=intx^2e^(x-1)dx=1/eintx^2e^xdx#

#intuv'=uv-intu'v#

Here,

#u=x^2#, #=>#, #u'=2x#

#v'=e^x#, #=>#, #v=e^x#

The integral is

#I=1/e(x^2e^x-2intxe^xdx)#

Apply the integration by parts once more

#u=x#, #=>#, #u'=1#

#v'=e^x#, #=>#, #v=e^x#

Therefore,

#intxe^xdx=xe^x-inte^xdx=xe^x-e^x#

Finally,

#I=1/e(x^2e^x-2(xe^x-e^x))+C#

#=e^x/e(x^2-2x+2)+C#

#=e^(x-1)(x^2-2x+2)+C#