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

[Sparky_]

Greetings -

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

Here's the integral:

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

I've tried some algebra and substitutions:

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

= u=e^x
= du=e^x dx

= \int -\frac{u}{(1+u)}du

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

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

= \int -\frac{1}{e^{-2x} +e^{-x}}dx

 

[Rainbow Child]

Split it in two partial fractions

\frac{u}{1+u}=\frac{u+1-1}{1+u}=1-\frac{1}{1+u}

 

[Sparky_]

Thanks - I'm still trying to hack my way around to correctly post the problem - I'm trying to find examples of this stuff, I can't quite get the problem to post correctly.

 

[Rainbow Child]

Check this

https://www.physicsforums.com/showthread.php?t=8997

 

[Sparky_]

So my integral is
\int du -\int \frac{1}{1+u} du

I know there will be a ln() involved eventually
?

 

[Rainbow Child]

Yes!
Which is equal to what?

 

[Sparky_]

Here's what I'm getting - can you confirm?

-e^x + ln|1+e^x|

 

[Rainbow Child]

Correct!

Only you forgot an overall minus sign, one from the inital integral!

 

[Sparky_]

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.

 

[Rainbow Child]

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!

 

[rocomath]

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.

 

[Sparky_]

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 -

 

[torquerotates]

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)

 

[Gib Z]

You used partial fractions.