How do you integrate #int xe^(x/2) dx#?

1 Answer
Jim H
Aug 23, 2015

Use integration by parts with #u = x# and #dv = e^(x/2)dx#.

Explanation:

#int xe^(x/2) dx#

Let #u = x# #" "#and #" "# #dv = e^(x/2)dx#

So #du = dx# #" "# and #" "# #v = 2e^(x/2)# #" "#(use substitution #w = x/2#)

#int xe^(x/2) dx = (x)(2e^(x/2)) - int 2e^(x/2) dx#

# = 2xe^(x/2) - 2 int e^(x/2) dx# #" "#(again use substitution #w = x/2#)

#= 2xe^(x/2) - 4e^(x/2) +C#

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