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Simplifying the result of integration

 
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Jul24-12, 02:03 PM   #1
LoA
 

Simplifying the result of integration


1. The problem statement, all variables and given/known data

The original problem is [itex] \int \, \frac {xe^{2x}}{(1+2x)^2} dx [/itex]. I utilized integration by parts to get:

[itex] -\frac {xe^{2x}}{2(1+2x)} \, + \, \frac {1}{4}e^{2x} \, + \, C [/itex]

which I know is correct. However, I am told by the book that this may also be expressed as:

[itex] \frac {e^{2x}}{4(2x+1)} + C [/itex]


It is a failing of my algebra skills that I am unable to make this translation. I have banged my head against this problem for close to an hour now. The truly frustrating thing is that it was very apparent to me yesterday (I was redoing the problem just to warm up for more i.b.p. problems), and I quickly wrote it down, but cannot seem to 'see' it today. Any help would be much appreciated.

2. Relevant equations

none

3. The attempt at a solution

I have tried expanding the denominator of the first term and factoring out the common [itex] \frac{1}{2} e^{2x} [/itex] but then I get stuck. In particular, I can't figure out how to deal with the x in the numerator and the negative sign. I think I'm pretty clearly forgetting/misapplying some algebraic rules, so the first thing in the way of suggestive assistance that might help would be just a reference to the relevant rules/ideas. I think that given such a push in the right direction I can figure this out on my own.

Additionally, if anyone could suggest some resources for practice with algebra of this level of difficulty I'd greatly appreciate it.
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Jul24-12, 02:24 PM   #2
 
Quote by LoA View Post
I have tried expanding the denominator of the first term
Not necessary.

Quote by LoA View Post
factoring out the common [itex] \frac{1}{2} e^{2x} [/itex]
Good start:) What does it look like after factoring out the term?

Quote by LoA View Post
but then I get stuck. In particular, I can't figure out how to deal with the x in the numerator and the negative sign.
It just means -x, or (-1)x. And I'm sure you know -b+a=a-b.
Do you remember how to add/subtract fractions with different denominators?
Jul24-12, 02:24 PM   #3
 
Mentor
Quote by LoA View Post
1. The problem statement, all variables and given/known data

The original problem is [itex] \int \, \frac {xe^{2x}}{(1+2x)^2} dx [/itex]. I utilized integration by parts to get:

[itex] -\frac {xe^{2x}}{2(1+2x)} \, + \, \frac {1}{4}e^{2x} \, + \, C [/itex]

which I know is correct. However, I am told by the book that this may also be expressed as:

[itex] \frac {e^{2x}}{4(2x+1)} + C [/itex]


It is a failing of my algebra skills that I am unable to make this translation. I have banged my head against this problem for close to an hour now. The truly frustrating thing is that it was very apparent to me yesterday (I was redoing the problem just to warm up for more i.b.p. problems), and I quickly wrote it down, but cannot seem to 'see' it today. Any help would be much appreciated.

2. Relevant equations

none

3. The attempt at a solution

I have tried expanding the denominator of the first term and factoring out the common [itex] \frac{1}{2} e^{2x} [/itex] but then I get stuck. In particular, I can't figure out how to deal with the x in the numerator and the negative sign. I think I'm pretty clearly forgetting/misapplying some algebraic rules, so the first thing in the way of suggestive assistance that might help would be just a reference to the relevant rules/ideas. I think that given such a push in the right direction I can figure this out on my own.

Additionally, if anyone could suggest some resources for practice with algebra of this level of difficulty I'd greatly appreciate it.
Multiply the first expression by 2/2, and multiply the second one by (1 + 2x)/(1 + 2x). That will give you two expressions with the same denominator, so you can combine them.
Jul24-12, 03:04 PM   #4
LoA
 

Simplifying the result of integration


Awesome, thank you, I'm really not sure why that didn't click. I think I frequently miss the forest for the trees

BTW, is the notion of a "simplified" related to how many operators have scope over the whole expression? i.e., here it went from two applications of "+" to just one. Just what is "simple" has never really been explained to me my satisfaction. Anyway, that's sort of asinine, but thanks again for your quick response and help with my problem!
Jul25-12, 01:28 PM   #5
 
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Quote by LoA View Post
BTW, is the notion of a "simplified" related to how many operators have scope over the whole expression? i.e., here it went from two applications of "+" to just one. Just what is "simple" has never really been explained to me my satisfaction. Anyway, that's sort of asinine, but thanks again for your quick response and help with my problem!
I don't know that there's a whole lot of agreement on what "simplified" means. In my experience it's somewhat subjective. However, most people would agree that the simplified form of 2x2 - 3x2 + 3 - 2x - 18 + 2x2 is x2 - 2x - 15.

Where they might not agree is whether the factored form is simpler than the form I show.
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