How can I integrate this expression?

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Discussion Overview

The discussion revolves around the integration of the expression \(\frac{e^{2x}}{e^{2x}+{3e^x}+2}\). Participants explore various methods for solving the integral, including substitution, partial fraction decomposition, and long division, while expressing uncertainty about their approaches and the correctness of their solutions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant attempted to factor the denominator and use partial fraction decomposition but expressed confusion due to a lack of examples in their textbook.
  • Another participant mentioned trying long division, which resulted in a simple term that did not match the expected answer.
  • A different approach involved substituting \(t = e^x\) and transforming the integral, suggesting further steps to simplify the expression.
  • One participant pointed out that using \(u = e^x\) allows for a different perspective on the integral, leading to a partial fractions setup that could be integrated.
  • A later reply indicated that a participant successfully solved the integral using a reference book, detailing their steps including long division and partial fraction decomposition, ultimately arriving at a simplified logarithmic expression.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to integrate the expression, with multiple approaches and some uncertainty about the correctness of their individual solutions.

Contextual Notes

Some participants express uncertainty about the integration techniques and the application of substitutions, indicating that they are still refining their understanding of the problem.

Who May Find This Useful

Students and individuals interested in integration techniques, particularly those struggling with similar expressions or looking for various methods to approach integration problems.

Pythagorean
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[tex]\frac{e^{2x}}{e^{2x}+{3e^x}+2}[/tex]

I tried factoring the bottom to ([tex]{e^x+2}[/tex])([tex]{e^x+1}[/tex]) and using PFDs

and I've also tried [tex]u=2x[/tex] and [tex]u=e^x[/tex]

We haven't covered e operations in class and th book gives no examples. I assumed it would just be simple enough to do (A + B)e^x.

The book's answer is [tex]ln \frac{({e^x}+2)^2}{e^x+1}[/tex]

I've filled up six pages with this problem, and I'm so close. is there something I'm forgetting from back in the day?
 
Last edited:
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i've also tried long division, but that yields a 1, which integrates into an X, which does not comply with the answer :/
 
let t=e^x, means 1-(3t+2)/(t^2+3t+2), let u=3t+2, take dt/du, then t^2=(u-2)^2/9, substitute this into the previous integrte, bring the denominator of the just-obtained integration to the form 1/(a^2+ or - b^2) and use known formula of this integration to get the wanted result, of course there are still some steps to make things neater. that is where i have come, have no time to go past the rest, you have to do that homework yourself. and hope this helps.
 
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Kroneker gave a good answer but I'll say my own piece: you said you tried u= ex. If you did then you found that du= exdx. You may have decided that that wouldn't work because you didn't have an "ex". Actually, you do! You have e2x= (ex)2= exex.
Letting u= ex the numerator is exexdx= u du while the denominator becomes u2+ 3u+ 2= (u+2)(u+1). The integral becomes
integral of udu/((u+2)(u+1)). That can be done by "partial fractions".
u/((u+2)(u+1))= A/(u+2)+ B/(u+1). Multiplying both sides by (u+2)(u+1), we get
u= A(u+1)+ B(u+2). Taking u= -1, that is -1= B. Taking u= -2, -2= -A so A= 1.
The integral is the same as 1/(u+2)- 1/(u+1). Can you integrat that?
 
wow, man, I finally figured this one out. Thank you everyone for your help. I actually solved this using a book in the reference section of the library to get how to integrate e^x.

The processes are a) long division, b) partial fraction decomposition (as suggest by Halls of Ivy), c) integration of terms, and finally d) arrangement of natural logs

it came out to be something like x-2x+2ln (x^2+2) + x - ln (x^2+1)

which simplifies to 2ln (x^2+2) - ln (x^2+1) which yields the answer from the first post when arranged properly for logs
 

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