# Integrals of e.

1. Sep 6, 2007

### niyati

I need to find the integral of (1/(1+e^x)) dx.

...I don't think I can use u substitution, but well, I'm basically open to suggestions.

2. Sep 6, 2007

### neutrino

It should be solvable with substitution. Hint: u = 1+e^x

3. Sep 6, 2007

### niyati

but, then wouldn't du = e^x dx? That would require a (1/e^x) outside the integral to balance it out, but my Cal I teacher (I'm now in Cal II with a different teacher) told me that I couldn't do this.

4. Sep 6, 2007

### neutrino

Write x in terms of u, and then take the differential.

5. Sep 6, 2007

### JonF

let u = 1 + e^x
then we know du = (e^x)dx
and e^x = u -1

is there something we can multiply both the numerator and the denominator by to make this substitution work? Hint: this will turn it into a partial fraction decomposition problem.

6. Sep 6, 2007

### d_leet

Multiply by 1. (1=e-x/e-x)

7. Sep 6, 2007

### HallsofIvy

Staff Emeritus
That multiplication isn't really necessary. JonF's first point was sufficient: if u= 1+ ex, then ex= u- 1. du= exdx= (u-1)dx so dx= du/(u-1).
$$\int\frac{dx}{1+e^x}= \int\frac{du}{(u-1)(u)}$$
which can be done by partial fractions.

Last edited: Sep 6, 2007
8. Sep 6, 2007

### PowerIso

Since i'm clumsy at partial fractions, I find multiplying by 1 to be a bit easier.

9. Sep 6, 2007

### bob1182006

Edit: woops sorry just saw d_leet mentioned this method already. But if you have a fraction with some e^x or e^-x then if you cant do u substitution, multiply by (e^-x)/(e^-x) or (e^x)/(e^x) and then do a nice u substitution.

another method:

multiply the top and bottom by e^-x to get:

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

which you can use the substitution u=e^-x +1 then you have du = e^-x dx at the top which reduces the problem down to a simple integral

Last edited: Sep 6, 2007
10. Sep 7, 2007

### VietDao29

Multiplying both numerator, and denominator by ex, or, making a substitution, is pretty much the same. After that, you have to do Partial Fraction it. You cannot avoid it. :)

11. Sep 7, 2007

### neutrino

I could waste time by going the trig-way.

12. Sep 7, 2007

### PowerIso

Really? I ended up with du/u which doesn't seem like I need to do partial fractions.

13. Sep 8, 2007

### JonF

Then I’m pretty sure you did something wrong, what substitution did you use?

14. Sep 8, 2007

### PowerIso

I used e^-x + 1 for u then du= -e^-x dx so since that is on the top already with just a different constant you can get -du/u.

15. Sep 8, 2007

### JonF

It’s the integral of: (1/(1+e^x)), not (1/(1+e^-x)). That "-" makes a huge difference. If you use that substitution you can’t substitute the bottom.

16. Sep 8, 2007

### HallsofIvy

Staff Emeritus
Nor was the integrand
$$\frac{e^{-x}dx}{1+ e^{-x}}$$
as PowerIso seems to think!

17. Sep 8, 2007

### bob1182006

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

which doing the substitution u=e^(-x)+1 gives you -du/u.

18. Sep 8, 2007

### camilus

I dont even think partial fractions are necessary. u(u-1)=u^2-u

so

$$\int\frac{dx}{1+e^x}= \int\frac{du}{(u-1)(u)} = \int{1 \over u^2-u}du = ln(u^2-u) + C = ln(e^{2x} + e^x) + C$$

19. Sep 8, 2007

### PowerIso

Well, here is what I did. I multiplied by e^-x/e^-x and I got e^-x/(e^-x + 1) , so i'm still fairly certain it works.

20. Sep 8, 2007

### PowerIso

Yes that is what I did :).