# 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.