Integration Problem: Solving an Equation with e

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Homework Help Overview

The discussion revolves around an integration problem involving the expression \(\int \frac{2e^{2t} \, dt}{e^{4t}+12e^{2t}+32}\). Participants are exploring methods to simplify and solve the integral, particularly focusing on the use of partial fractions and substitutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to factor the denominator and express the integral in terms of partial fractions but encounters difficulties with the resulting expressions. Some participants suggest using a substitution \(u = e^{2t}\) to facilitate the integration process, while others express confusion about the factorization of the terms involved.

Discussion Status

The discussion is active, with participants providing various suggestions for approaching the problem. There is an acknowledgment of the need for further integration steps, and some guidance has been offered regarding the use of substitutions and partial fractions. However, there is no explicit consensus on the best path forward.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the methods they can employ. There is also uncertainty regarding the factorization of certain expressions, which is a point of contention in the discussion.

Melawrghk
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Homework Statement


[tex]\int \frac{2e^{2t}*dt}{e^{4t}+12e^{2t}+32}[/tex]

The Attempt at a Solution


I factored the denominator, getting (e2t+4)(e2t+8), from that I was able to express the original equation in terms of two fractions:
[tex]\int \frac{A}{e^{2t}+4}[/tex]+[tex]\frac{B}{e^{2t}+8}[/tex]

I then found A to be -2 and B=4.

But when I put those back into the original equations, I still have integrals that have an 'e' on the bottom and I don't know what to do next...

Any help is appreciated.
 
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Put u=e2t, and then split into partial fractions.
 
More partial fractions? But I can't. u+4 doesn't factor...?
 
Melawrghk said:
More partial fractions? But I can't. u+4 doesn't factor...?

[tex] \int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt[/tex]

u=e^2t => du/2 =e^2t dt

[tex]\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt \equiv \frac{1}{2} \int \frac{1}{u^2+12u+32} du[/tex]

Now do partial fractions.
 
I already did. It's in my procedure...
 
Then you'd simply have to integrate

[tex]\frac{1}{8} \int (\frac{1}{u+4} - \frac{1}{u+8})du[/tex]
 

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