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Integration problem

  • #1
The question is to evaluate the integral in the attachment. My idea was to use partial fraction decomposition and so my new integral looked like:

A/((e^x)+1) + B/((e^x)+5) where I got A=-2 and B=-10

When I integrate this, I get:

-2ln((e^x)+1) - 10ln((e^x)+5)

However, this is not the answer.
 

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Answers and Replies

  • #2
Dick
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I can tell you right away that the integral of 1/(e^x+1) is not ln(e^x+1). Differentiate the latter (remembering the chain rule) to see why not. To do this correctly substitute u=(e^x+1).
 
Last edited:
  • #3
2,063
2
I cannot see the integral till the attatchment it is approved but how did you manage to integrate something of the form

[tex]\frac{A}{e^x+a}[/tex] to

[tex]A\ln{(e^x+a)}[/tex]

?

One way to check if your integration is done correctly is to differentiate the function you arrived at and see if it matches the earlier function.
 
  • #4
For ∫-2/((e^x)+1) dx, I choose u=(e^x)+1 so du=(e^x) dx so I get:

-2 ∫ du / u(e^x) where (e^x)= u-1 therefore -2 ∫ du / (u^2)-u

Is this correct?
 
  • #5
Dick
Science Advisor
Homework Helper
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Yes. So you see you need another partial fractions thing to solve that. Because you WOULDN'T just write log(u^2-u), would you?
 
  • #6
No I would not!
 
  • #7
Got it... thanks a lot!
 

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