How to integrate #int xe^(x/2)dx# by parts?

1 Answer
Jul 20, 2018

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

Explanation:

The integration by parts is

#intuv'dx=uv-intu'vdx#

Here,

#u=x#, #=>#, #u'=1#

#v'=e^(x/2)#, #=>#, #v=2e^(x/2)#

Therefore,

The integral is

#I=intxe^(x/2)dx=2xe^(x/2)-2inte^(x/2)dx#

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