Integrating (1/(1+e^x)) dx: Challenges and Solutions

[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.

 

[neutrino]

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

 

[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.

 

[neutrino]

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

 

[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.

 

[d_leet]

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

 

[HallsofIvy]

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

 

[PowerIso]

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

 

[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 can't 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:

[tex]\int\frac{e^{-x} dx}{e^{-x}+1}[/tex]

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

 

[VietDao29]

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

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

 

[neutrino]

After that, you have to do Partial Fraction it. You cannot avoid it. :)

I could waste time by going the trig-way.

 

[PowerIso]

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

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

 

[JonF]

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

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

 

[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.

 

[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.

 

[HallsofIvy]

Nor was the integrand
[tex]\frac{e^{-x}dx}{1+ e^{-x}}[/tex]
as PowerIso seems to think!

 

[bob1182006]

no what PowerIso did was this I think:

[tex]\int\frac{dx}{1+e^x} * \frac{e^{-x}}{e^{-x}} = \int\frac{e^{-x} dx}{e^{-x}+1}[/tex]

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

 

[camilus]

I don't even think partial fractions are necessary. u(u-1)=u^2-u

so

[tex]\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[/tex]

 

[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.

 

[PowerIso]

no what PowerIso did was this I think:

[tex]\int\frac{dx}{1+e^x} * \frac{e^{-x}}{e^{-x}} = \int\frac{e^{-x} dx}{e^{-x}+1}[/tex]

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

Yes that is what I did :).

 

[GoldPheonix]

How on Earth do you do partial fractions on exponential functions?

 

[bob1182006]

you don't do it directly you substitute to get some polynomials.

@ camilus: that integration is wrong you don't have the derivative of (u^2-u) on the top which is needed in order to get ln(u^2-1) + C as the answer so you would have to do partial fractions.

 

[GoldPheonix]

I don't even think partial fractions are necessary. u(u-1)=u^2-u

so

[tex]\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[/tex]

Nope, doesn't work out...

The logorithmic integration rules do not work that way. The integral of "1/(x+x^2)" =/= ln(x+x^2)

You'd have to start doing trigonometric sub/partial fractions after this point.

 

[JonF]

no what PowerIso did was this I think:

[tex]\int\frac{dx}{1+e^x} * \frac{e^{-x}}{e^{-x}} = \int\frac{e^{-x} dx}{e^{-x}+1}[/tex]

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

slick ;)

 

[camilus]

[tex]\int \limits_0^1 \frac{dx}{1+e^x} \approx 0.379885493[/tex]

[tex]\int \limits_2^{e+1} {1 \over u^2-u}du \approx 0.379885493[/tex]

most of yall need to retake calc 1, apparently everyone disagrees with the correct answer...

 

[jacobrhcp]

you're missing some calculus there, the derivative of ln(u^2-u) is not your integral.

 

[camilus]

not ln(u^2+u), ln(u^2-u)...

 

[jacobrhcp]

yea I meant that >_> sorry

 

[dextercioby]

And it's still not true. [itex] \frac{d}{du}\ln\left(u^{2}-u\right) \neq \frac{1}{u^{2}-u} [/itex].

 

[PowerIso]

Hmm d/dxln(u^2-u) is (1/u^2 - u) * 2u

I'm pretty sure I did the derivative right, so camulis where did I error?

 

[dextercioby]

Are you differentiating [itex] \ln(u^2 -u) [/itex] wrt "x" or to "u" ?

 

[PowerIso]

opps I meant to type d/du

 

[Hurkyl]

[tex]\int{1 \over u^2-u}du = ln(u^2-u) + C[/tex]

Why would you think that?



p.s. have you looked at your private messages?

 

[camilus]

http://i159.photobucket.com/albums/t121/camilus23/etothexproblem.jpg


Summary:

[tex]\int {1 \over e^x +1}dx = ln \left|{e^x \over e^x+1}\right| + C[/tex]

 

[asd1249jf]

Lol jesus how many mathematicians does it take to solve an exponential integral problem?

I'm not calling anyone stupid, but just found it funny ya'll are being cooperative to solve this