Integral Help: Solving -e^x/(e^x+1)

In summary, the conversation discusses a problem involving an integral and different methods for solving it. The integral is transformed using algebra and substitutions, and it is suggested to use partial fractions or another substitution to solve it. The correct solution is found to be -e^x + ln|1+e^x|, with the suggestion to use polynomial division next time.
  • #1
Sparky_
227
5
Greetings -

This integral is part of a larger problem I'm working on - I'm stuck (I Think)

Here's the integral:

[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]

I've tried some algebra and substitutions:

[tex] = \int -\frac{e^{2x}}{(1+e^x)}dx[/tex]

[tex] = u=e^x[/tex]
[tex] = du=e^x dx[/tex]

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

Is this correct? Is this a standard form for something? - looking like ln()?

OR
could go:
[tex] = \int -\frac{e^{-x}}{e^{-3x}(1+e^x)}dx[/tex]

[tex] = \int -\frac{1}{e^{-2x} +e^{-x}}dx[/tex]
 
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  • #2
Split it in two partial fractions

[tex]\frac{u}{1+u}=\frac{u+1-1}{1+u}=1-\frac{1}{1+u}[/tex]
 
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  • #3
Thanks - I'm still trying to hack my way around to correctly post the problem - I'm trying to find examples of this [tex] stuff, I can't quite get the problem to post correctly.
 
  • #5
So my integral is
[tex] \int du -\int \frac{1}{1+u} du[/tex]

I know there will be a ln() involved eventually
?
 
  • #6
Yes!
Which is equal to what?
 
  • #7
Here's what I'm getting - can you confirm?

[tex] -e^x + ln|1+e^x| [/tex]
 
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  • #8
Correct!

Only you forgot an overall minus sign, one from the inital integral!
 
  • #9
oops

thanks

By the way, can you suggest any other method that would have solved this integral?

I have not used partial fractions in a long time, if you had not suggested it, I would not have gone there.

The integral looked simple enough for me to "pencil and paper" it and impress my friends without the need for a CRC handbook or other aid - I was wrong.
 
  • #10
But this is the more efficient way! To try to transform integrals with log's, sin's, cos's, exp's to rational form, and then apply partial fractions!

At least that' s the way I think! :smile:
 
  • #11
Well you could have used polynomial division if you had not thought of adding +/- 1 to the numerator so that you could split it into two.
 
  • #12
I guess I need to knock off the rust and practice up a little bit so the next time it's more obvious.

Thank you all for helping with this -
 
  • #13
Instead of using partial fractions. Just use another substitution. Use say t=1+u. This implies that u=t-1. So you have integral of (t-1)/t. And keep in mind that dt=du so this integral is relatively easy. Because its just the integral of 1-(1/t). which is t-ln(t). Just back sub for the t. so 1+u-ln(1+u)
 
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  • #14
You used partial fractions.
 

1. What is "Integral Help"?

Integral Help is a mathematical concept that involves finding the area under a curve on a graph. It is also known as integration and is the inverse operation of differentiation.

2. What does "Solving -e^x/(e^x+1)" mean?

"Solving -e^x/(e^x+1)" refers to finding the integral of the function -e^x/(e^x+1). This means finding the function whose derivative is -e^x/(e^x+1).

3. What is the purpose of solving this integral?

The purpose of solving this integral is to find the area under the curve of the function -e^x/(e^x+1). This can be useful in various real-life applications, such as calculating the work done by a varying force.

4. What are the steps to solve this integral?

The steps to solve this integral involve using integration by parts or substitution methods to manipulate the function into a form that can be easily integrated. Then, the integral can be evaluated using integration rules and techniques.

5. Are there any special cases or exceptions when solving this integral?

Yes, there are some special cases and exceptions when solving this integral. For example, if the limits of integration are infinite, the integral may not converge and therefore cannot be solved. Additionally, certain functions may require special techniques or rules to be integrated.

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