How do you find the integral of #xe^(x/2)dx #?

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Answer 1 · 2015-05-26T01:01:14

By part

#u = x#
#du = 1#

#dv = e^(1/2x)#

#v = 2e^(1/2x)#

#=>2([xe^(1/2x)]-inte^(1/2x)dx)+C#

#=>2([xe^(1/2x)]-2[e^(1/2x)])+C#

#=>2e^(1/2x)(x-2)+C#