How do you integrate #int re^(r/2)# by integration by parts method?

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How do you integrate #int re^(r/2)# by integration by parts method?

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Answer 1 · 2016-10-20T12:50:05

#intre^(r/2)dr=(2r-4)e^(r/2)+C#

Explanation:

Let #u=r# then #u'=1#
and #v'=e^(r/2)# then #v=2e^(r/2)#
#intuv'=uv-intu'v#
so #intre^(r/2)dr=2re^(r/2)-int2e^(r/2)dr#
#=2re^(r/2)-2*2e^(r/2)#
#=2re^(r/2)-4e^(r/2)+C#
#=(2r-4)e^(r/2)+C#

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How do you integrate #int re^(r/2)# by integration by parts method?

1 Answer

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