Integrating Partial Fractions: Evaluating Difficult Integral

[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.

 

[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).

 

[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.

 

[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?

 

[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?

 

[cmantzioros]

No I would not!

 

[cmantzioros]

Got it... thanks a lot!