Integration problem

1. Apr 7, 2007

cmantzioros

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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2. Apr 7, 2007

Dick

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: Apr 7, 2007
3. Apr 7, 2007

neutrino

I cannot see the integral till the attatchment it is approved but how did you manage to integrate something of the form

$$\frac{A}{e^x+a}$$ to

$$A\ln{(e^x+a)}$$

?

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. Apr 7, 2007

cmantzioros

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. Apr 7, 2007

Dick

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. Apr 7, 2007

cmantzioros

No I would not!

7. Apr 7, 2007

cmantzioros

Got it... thanks a lot!

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