What is the correct approach to solving this integral?

[QuarkCharmer]

Homework Statement

\int \frac{e^{2x}}{e^{2x}+3e^{x}+2}dx

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

\int \frac{e^{2x}}{e^{2x}+3e^{x}+2}dx

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

e^{2x} = e^{2x}(A+B)+e^{x}(2A+B)

A + B = 1
2A + B = 0
A=-1 and B=2

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

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

??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?

 

[SammyS]

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".

\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}

\displaystyle =1-\frac{3e^{x}+2}{e^{2x}+3e^{x}+2}​

Now use partial fractions.

 

[Dick]

Homework Statement

\int \frac{e^{2x}}{e^{2x}+3e^{x}+2}dx

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

\int \frac{e^{2x}}{e^{2x}+3e^{x}+2}dx

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

e^{2x} = e^{2x}(A+B)+e^{x}(2A+B)

A + B = 1
2A + B = 0
A=-1 and B=2

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

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

??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.

 

[QuarkCharmer]

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:
ln(\frac{(e^{x}+2)^{2}}{(e^{x}+1)})+C

But I am not seeing how you can use the properties of log to equate that to my solution?

 

[SammyS]

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 !

 

[Dick]

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:
ln(\frac{(e^{x}+2)^{2}}{(e^{x}+1)})+C

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?

 

[QuarkCharmer]

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:
ln(a) - 2ln(b) = ...

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.

 

[Dick]

Yeah I know those, but I don't have a condition like that (at least it's not apparent to me).
I have:
ln(a) - 2ln(b) = ...

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.

 

[Trevor Vadas]

I would argue for your points.
what you had and the answer are equivilent.

-ln(a) + 2ln(b) = ln(a^{-1}) + ln(b^{2})
= ln((a^{-1})(b^{2})) = ln (^{b^{2}}_{a})