Integral kicking my butt

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In summary, the author is trying to solve a homework equation but does not understand what he is doing. He missed one question on a quiz but it looks right to him. He has gone over it a dozen times. The fraction part of the equation is:
  • #1
QuarkCharmer
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Homework Statement


[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.

Homework Equations



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?
 
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  • #2
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.
 
  • #3
QuarkCharmer said:

Homework Statement


[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.

Homework Equations



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?

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.
 
  • #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?
 
  • #5
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 !
 
  • #6
QuarkCharmer said:
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?

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?
 
  • #7
Dick said:
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?

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.
 
  • #8
QuarkCharmer said:
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.

Right, use it to put the exponent back up.
 
  • #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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