Integration Problem: Solving an Equation with e

  • Thread starter Melawrghk
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In summary, the original integral can be solved by first factoring the denominator and then expressing it as a sum of two fractions. After finding the values for A and B, the integral can be rewritten in terms of u=e^2t and solved through partial fractions. This leads to a final integral that can be easily integrated.
  • #1
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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  • #2
Put u=e2t, and then split into partial fractions.
 
  • #3
More partial fractions? But I can't. u+4 doesn't factor...?
 
  • #4
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.
 
  • #5
I already did. It's in my procedure...
 
  • #6
Then you'd simply have to integrate

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

Related to Integration Problem: Solving an Equation with e

1. What is the general process for solving an equation with e?

The general process for solving an equation with e is to first isolate the term containing e on one side of the equation. Then, take the natural logarithm of both sides to eliminate the e. Finally, solve for the remaining variable.

2. How do I know when to use the natural logarithm when solving an equation with e?

You should use the natural logarithm when the equation contains e as an exponent or when the equation can be rewritten to have e as an exponent. This is because taking the natural logarithm will eliminate the e, making it easier to solve for the variable.

3. Can an equation with e have multiple solutions?

Yes, an equation with e can have multiple solutions. This is because the natural logarithm is a multivalued function, meaning it can have multiple solutions for a given input. Therefore, when solving an equation with e, you should always check for extraneous solutions.

4. Can e be solved for in an equation?

No, e cannot be solved for in an equation. This is because e is a constant, representing a specific numerical value (approximately 2.71828). It is not a variable that can be solved for.

5. Are there any specific rules for solving an equation with e?

Yes, there are a few rules to keep in mind when solving an equation with e. First, the natural logarithm of a negative number is undefined, so any solutions that result in a negative number under the natural logarithm should be discarded. Additionally, when taking the natural logarithm of both sides, the result should be simplified to the form "x = ...".

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