7.1 Integration By Parts/4: Difference between revisions

Revision as of 02:59, 29 November 2022

$\int (xe^{-x})dx$

$u=x\qquad dv=e^{-x}$

$du=dx\qquad v=-e^{-x}$

$\int (xe^{-x})dx=x(-e^{-x})-\int (-e^{-x})=-xe^{-x}-\int (-e^{-x})dx$

$-xe^{-x}+\int (xe^{-x})dx=$

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	<math>-xe^{-x}+\int (xe^{-x})dx =		<math>-xe^{-x}+\int (xe^{-x})dx =
	<\math>		</math>