How do you integrate #int xe^(x/2)# using integration by parts?

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Answer 1 · 2018-04-20T18:40:26

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

Make the following selections:

#u=x#

#du=dx#

#dv=e^(x/2)dx#

#v=inte^(x/2)dx=2e^(x/2)#

Then

#uv-intvdu=2xe^(x/2)-2inte^(x/2)dx=2xe^(x/2)-4e^(x/2)+C#

Factoring out the exponential, we obtain

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