Partial Fractions: Solving ∫ 5e^2x / (25e^2x - 20e^x +4) dx

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Homework Help Overview

The problem involves evaluating the integral ∫ 5e^2x / (25e^2x - 20e^x + 4) dx, which presents challenges due to the exponential nature of the denominator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to handle the integral given the exponential denominator and questions what form the numerator should take. Some participants suggest using a substitution, specifically y = e^x, to simplify the problem.

Discussion Status

Participants are exploring the substitution method and discussing how it could facilitate the computation of partial fractions. There is acknowledgment of the technique's relevance to future topics in the course.

Contextual Notes

The discussion includes considerations about the structure of the integral and the potential need for variable substitution to make the problem more manageable. There is also mention of the original poster's unfamiliarity with the forum's math formatting.

maff is tuff
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Homework Statement



∫ 5e^2x / (25e^2x - 20e^x +4) dx

Homework Equations






The Attempt at a Solution



The attempt at the the solution is in the attachment below. I am stuck at that step. Pretty much can anyone tell me what 'something' should be. I know if the denominator were a reducible quadratic then the numerator would be linear (Ax + B). But what if the denominator is exponential like in my problem?

 

Attachments

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maff is tuff said:
I know if the denominator were a reducible quadratic then the numerator would be linear (Ax + B). But what if the denominator is exponential like in my problem?

It will be something like [tex]A e^x +B[/tex].

It might be easiest for you to perform a change of variable [tex]y=e^x[/tex] to reduce the problem to a more familiar one.
 
So everywhere I see an e^x I put a y? Then go back in later and put the e^x's back?
 
maff is tuff said:
So everywhere I see an e^x I put a y? Then go back in later and put the e^x's back?

You could do that in order to make the partial fraction computation easier. You could also go further and use

[tex]dy = \frac{dy}{dx} dx[/tex]

to solve for [tex]dx[/tex] in terms of [tex]dy[/tex], which would allow you to rewrite the integral itself as an integral over the new variable [tex]y[/tex]. You should see this technique come up quite a bit later in your course.

Just to illustrate this with an example, suppose we have

[tex]I = \int x^3 dx = \frac{x^4}{4} + c.[/tex]

If we set [tex]y = x^2[/tex], we see that [tex]dy = 2 x dx[/tex], so we can rewrite

[tex]I = \int \frac{y ~dy}{2} = \frac{y^2}{4} + c = \frac{x^4}{4} + c.[/tex]

The reason why the change of variables works is tied to the chain rule for derivatives.
 
Thanks a lot for your help and for taking the time to write out such a detailed explanation. Also, how do you type your math symbols like that? Sorry, I'm relatively new to this site and every time I type math symbols using the ones the forum gives me they don't look like that. Thanks again.
 
maff is tuff said:
Thanks a lot for your help and for taking the time to write out such a detailed explanation. Also, how do you type your math symbols like that? Sorry, I'm relatively new to this site and every time I type math symbols using the ones the forum gives me they don't look like that. Thanks again.

It's called LaTeX code and there's a sticky thread about it here: https://www.physicsforums.com/showthread.php?t=386951 You can also click on the formulas above to see the code that was used to make them.
 
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