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Integrate (xe^(2x))/(1+2x)^2

by autodidude
Tags: integrate, xe2x or 1
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autodidude
#1
Feb17-13, 12:04 AM
P: 333
1. The problem statement, all variables and given/known data
Integrate [tex]\frac{xe^{2x}}{(1+2x)^2}[/tex] with respect to x

Didn't get anywhere with integration by parts or substitution using u=xe^(2x)
A push in the right direction would be much appreciated.
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voko
#2
Feb17-13, 01:13 AM
Thanks
P: 5,794
Try v = 1 + 2x.
autodidude
#3
Feb17-13, 10:27 AM
P: 333
As a second substitution?

voko
#4
Feb17-13, 10:29 AM
Thanks
P: 5,794
Integrate (xe^(2x))/(1+2x)^2

No, just start with it.
autodidude
#5
Feb17-13, 02:41 PM
P: 333
Ok, I now have the following:

[tex]\frac{1}{4} \int \frac{(u-1)e^{(u-1)}{u^2}[/tex]
Karnage1993
#6
Feb17-13, 02:45 PM
P: 132
Quote Quote by autodidude View Post
Ok, I now have the following:

[tex]\frac{1}{4} \int \frac{(u-1)e^{(u-1)}{u^2}[/tex]
Allow me to fix that for you:

##\displaystyle \frac{1}{4} \int \frac{(u-1)e^{(u-1)}}{u^2} \ du##
voko
#7
Feb17-13, 02:50 PM
Thanks
P: 5,794
where is du?
ehild
#8
Feb17-13, 03:43 PM
HW Helper
Thanks
P: 10,607
Quote Quote by autodidude View Post
1. The problem statement, all variables and given/known data
Integrate [tex]\frac{xe^{2x}}{(1+2x)^2}[/tex] with respect to x

Didn't get anywhere with integration by parts or substitution using u=xe^(2x)
A push in the right direction would be much appreciated.
Integrate by parts

∫uv'dx=uv-∫u'vdx,

using u=xe2x and v'=1/(1+2x)2.

ehild
Karnage1993
#9
Feb17-13, 03:52 PM
P: 132
Quote Quote by ehild View Post
Integrate by parts

∫uv'dx=uv-∫u'vdx,

using u=xe2x and v'=1/(1+2x)2.

ehild
Parts requires u,v to be continuous.
voko
#10
Feb17-13, 04:05 PM
Thanks
P: 5,794
Now that we have reinstated du, observe that e^(u - 1) = (e^u)/e; the 1/e constant goes outside, and what's inside can be simplified into ((e^u)/u - (e^u)/u^2).


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