How can I integrate this expression?

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In summary, the conversation discusses different attempts at finding the integral of e^(2x) divided by e^(2x)+3e^x+2. The proposed solution involves using partial fractions and integrating using substitution and known formulas for integration. Ultimately, the answer is found to be 2ln(x^2+2) - ln(x^2+1).
  • #1
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?
 
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  • #2
i've also tried long division, but that yields a 1, which integrates into an X, which does not comply with the answer :/
 
  • #3
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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  • #4
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?
 
  • #5
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
 

1. What is the basic concept of Integration of e powers?

The concept of Integration of e powers involves finding the antiderivative or integral of a function that contains an e power. In other words, it is the process of finding a function whose derivative is the original function.

2. Why is the e power commonly used in integration?

The e power is commonly used in integration because it is the base of the natural logarithm function and has many useful properties that make it easier to integrate. Additionally, many real-life applications involve exponential growth or decay, which can be represented using the e power.

3. How do you integrate a function with an e power?

To integrate a function with an e power, you can use several techniques such as u-substitution, integration by parts, or using the properties of the e power. It is important to understand the rules and properties of integration as well as practice solving various types of integration problems.

4. What is the difference between indefinite and definite integration of e powers?

Indefinite integration of e powers involves finding a general antiderivative of a function, while definite integration involves finding the numerical value of the integral over a specific interval. In other words, indefinite integration gives a function, while definite integration gives a number.

5. Are there any common mistakes to avoid when integrating e powers?

Yes, some common mistakes when integrating e powers include forgetting to add the constant of integration, not applying the correct integration rules, and not simplifying the final answer. It is important to check your work and make sure it follows the proper steps of integration.

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