Integrate #int xe^(2x)dx# ?

2 Answers
Jul 10, 2018

This will require a single application of integration by parts.

We let #dv = e^(2x)dx# and #u = x#. Then #v= 1/2e^(2x)# and #du = dx#.

By integration by parts we have

#int u dv = uv - int v du#

#int xe^(2x) dx = 1/2xe^(2x) - int 1/2e^(2x) dx#

#int xe^(2x) dx = 1/2xe^(2x) - 1/4e^(2x) + C#

Hopefully this helps!

Jul 10, 2018

The answer is #=(2x-1)e^(2x)/4+C#

Explanation:

Perform an integration by parts

#intuv'dx=uv-intu'vdx#

Here,

The integral is

#I=intxe^(2x)dx#

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

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

Therefore, the integral is

#I=(xe^(2x))/2-1/2inte^(2x)dx#

#=(xe^(2x))/2-1/4e^(2x)+C#

#=(2x-1)e^(2x)/4+C#