How do you find the antiderivative of #((2x)e^(3x))#?

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Answer 1 · 2016-08-03T13:18:44

#1/9*(3x-1)*e^(3x)+C#.

Let, #I=int2xe^(3x)dx rArr I=2intxe^(3x)dx#.

To find #I#, we will use the following Rule of Integration by Parts :

#intuvdx=uintvdx-int{(du)/dxintvdx}dx#.

We take, #u=x, so, (du)/dx=1, &, v=e^(3x), so, intvdx=1/3e^(3x)#. So,

#I=x*1/3e^(3x)-int{1*1/3e^(3x)}dx#

#=x/3e^(3x)-1/3inte^(3x)dx#

#=x/3e^(3x)-1/3*1/3e^(3x)#

#:. I = 1/9*(3x-1)*e^(3x)+C#.