Simplifying the result of integration

In summary: In some circumstances factored form is simpler, but in others, like when the coefficients are irrational numbers, the factored form can be more complicated. So "simplified" can be a matter of opinion.
  • #1
LoA
13
0

Homework Statement



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.

Homework Equations



none

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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  • #2
LoA said:
I have tried expanding the denominator of the first term
Not necessary.

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

LoA said:
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?
 
  • #3
LoA said:

Homework Statement



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.

Homework Equations



none

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.
 
  • #4
Awesome, thank you, I'm really not sure why that didn't click. I think I frequently miss the forest for the trees :redface:

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!
 
  • #5
LoA said:
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.
 

1. What is integration and why is it important?

Integration is a mathematical process that involves finding the area under a curve. It is important because it allows us to calculate quantities such as displacement, velocity, and acceleration in physics, as well as determine probabilities in statistics.

2. Why do we need to simplify the result of integration?

Simplifying the result of integration helps us to better understand and interpret the information that we have calculated. It also allows us to easily compare and analyze different functions.

3. How do we simplify the result of integration?

The process of simplifying the result of integration involves using various integration techniques, such as substitution and integration by parts, as well as algebraic manipulation and trigonometric identities.

4. What are the common mistakes to avoid when simplifying the result of integration?

Some common mistakes to avoid include incorrectly applying integration techniques, forgetting to account for constant terms, and making algebraic errors. It is important to carefully check your work and practice regularly to avoid these mistakes.

5. Are there any shortcuts or tricks to simplify the result of integration?

There are certain common integrals and patterns that can be memorized to simplify the result of integration. Additionally, practicing regularly can help to develop a better understanding and intuition for integration, making it easier to simplify results. However, it is important to always double check your work and not rely solely on shortcuts.

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