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Integral kicking my butt

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{e^{2x}}{e^{2x}+3e^{x}+2}dx[/tex]

    I don't understand what I am doing wrong here. I missed this one question on a quiz but it looks right to me, I have went over it a dozen times.

    2. Relevant equations

    3. The attempt at a solution

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

    Fraction part:
    [tex]\frac{e^{2x}}{(e^{x}+1)(e^{x}+2)} = \frac{Ae^{x}}{e^{x}+1} + \frac{Be^{x}}{e^{x}+2}[/tex]

    [tex]e^{2x} = e^{2x}(A+B)+e^{x}(2A+B)[/tex]

    [itex]A + B = 1[/itex]
    [itex]2A + B = 0[/itex]
    [itex]A=-1[/itex] and [itex]B=2[/itex]

    Integral stuff:
    [tex]\int -\frac{e^{x}}{e^{x}+1} + \frac{2e^{x}}{e^{x}+2}dx[/tex]

    My solution:
    [tex]-ln(|e^{x}+1|) + 2ln(|e^{x}+2|) + C[/tex]

    ??


    Edit: Additionally:
    If one of the factors were (e^(2x)+1), does that count as a non-reducible polynomial and get A(e^x)+B as it's numerator instead of just plain A?
     
    Last edited: Oct 3, 2011
  2. jcsd
  3. Oct 3, 2011 #2

    SammyS

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    It's as if the numerator of the integrand is of the same degree as the denominator. Use long division, or the following trick to reduce the "improper fraction", to a "mixed fraction".

    [itex]\displaystyle \frac{e^{2x}}{e^{2x}+3e^{x}+2}=\frac{e^{2x}+3e^{x}+2}{e^{2x}+3e^{x}+2}-\frac{3e^{x}+2}{e^{2x}+3e^{x}+2}[/itex]
    [itex]\displaystyle =1-\frac{3e^{x}+2}{e^{2x}+3e^{x}+2}[/itex]​
    Now use partial fractions.
     
  4. Oct 3, 2011 #3

    Dick

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    I think you already have a correct solution there. It's a little unconventional, I think most people would substitute u=e^x first and then work with the polynomials in u. But I think you pulled it off.
     
  5. Oct 3, 2011 #4
    Oh yeah, a sub would have made that a bit less annoying. Is it still correct though that I did the partial fraction decomposition when the degree of the num/den were equal?

    For reference, the "correct" solution was:
    [itex]ln(\frac{(e^{x}+2)^{2}}{(e^{x}+1)})+C[/itex]

    But I am not seeing how you can use the properties of log to equate that to my solution?
     
  6. Oct 3, 2011 #5

    SammyS

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    QC,

    I didn't look at your complete solution. I looked at your title & just suggested the usual first step to take before doing partial fractions.

    Your solution is very interesting. I like it !!!
     
  7. Oct 3, 2011 #6

    Dick

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    You did a correct decomposition of the integral into two things you can integrate. I wouldn't exactly call it 'partial fractions' but it works. The only rules of log you need to show they are equal are things like log(a^b)=b*log(a) and log(a/b)=log(a)-log(b). You know those, right?
     
  8. Oct 3, 2011 #7
    Yeah I know those, but I don't have a condition like that (at least it's not apparent to me).
    I have:
    [itex]ln(a) - 2ln(b) = ...[/itex]

    I couldn't figure out how to get rid of that 2.

    Edit: Oh wait, you used that property to stick it back up into the exponent... That's where they got that squared term from. I never would have considered that lol. I just use that one to pull things down.
     
  9. Oct 3, 2011 #8

    Dick

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    Right, use it to put the exponent back up.
     
  10. Oct 3, 2011 #9
    I would argue for your points.
    what you had and the answer are equivilent.

    -ln(a) + 2ln(b) = ln(a[itex]^{-1}[/itex]) + ln(b[itex]^{2}[/itex])
    = ln((a[itex]^{-1}[/itex])(b[itex]^{2}[/itex])) = ln ([itex]^{b^{2}}_{a}[/itex])
     
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