How do you integrate #int xe^-x# by integration by parts method?

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Answer 1 · 2016-11-27T17:30:57

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

We have the integral #intxe^-xdx#. We want to apply integration by parts, which fits the form #intudv=uv-intdvu#. So for the given integral, let:

#{(u=x,=>,du=dx),(dv=e^-xdx,=>,v=-e^-x):}#

To go from #dv# to #v#, the integration will require a substitution. Think for #inte^-xdx#, let #t=-x#.

So:

#intxe^-xdx=uv-intudv=-xe^-x+inte^-xdx#

We've already done this integral:

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

#intxe^-xdx=-e^-x(x+1)+C#