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,

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

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

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

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