How do I evaluate #int(z +1) e^(3z) dz#?

1 Answer
Mind
Jan 28, 2015

#int (z+1)e^(3z) dz = int z*e^(3z)dz + int e^(3z)dz#

Second component is trivial
#int e^(3z) dz = int e^(3z)/3 d(3z) = e^(3z)/3 #

For the first component, we use integration by part:
Recall: Integration by part: #int u(dv) = uv - int v(du) #
let #u = z# and #dv = e^(3z)#.
If #dv = e^(3z)#, then #v = e^(3z)/3 #
(from second component)

Thus, #int z*e^(3z) dz = z*e^(3z)/3 - int e^(3z) dz #

Put the two component together:

#int (z+1)e^(3z) dz #
#= int z*e^(3z) dz + int e^(3z)dz #
#= [ z*e^(3z)/3 - int e^(3z) dz] + int e^(3z)dz #
#= z*e^(3z)/3 + C #

Hope this help!

Related questions
See all questions in Integration by Parts
Impact of this question
5130 views around the world
You can reuse this answer
Creative Commons License