Integration Question

1. Jan 3, 2008

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

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

Last edited: Jan 4, 2008
2. Jan 3, 2008

Rainbow Child

Split it in two partial fractions

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

Last edited: Jan 3, 2008
3. Jan 4, 2008

Thanks - I'm still trying to hack my way around to correctly post the problem - I'm trying to find examples of this $$stuff, I can't quite get the problem to post correctly. 4. Jan 4, 2008 Rainbow Child 5. Jan 4, 2008 Sparky_ So my integral is [tex] \int du -\int \frac{1}{1+u} du$$

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

6. Jan 4, 2008

Rainbow Child

Yes!
Which is equal to what?

7. Jan 4, 2008

Sparky_

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

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

Last edited: Jan 4, 2008
8. Jan 4, 2008

Rainbow Child

Correct!

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

9. Jan 4, 2008

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.

10. Jan 4, 2008

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!

11. Jan 4, 2008

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.

12. Jan 4, 2008

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 -

13. Jan 12, 2008

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)

Last edited: Jan 12, 2008
14. Jan 12, 2008

Gib Z

You used partial fractions.

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