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).
\int\frac{dx}{1+e^x}= \int\frac{du}{(u-1)(u)}
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:

\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

 

[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
\frac{e^{-x}dx}{1+ e^{-x}}
as PowerIso seems to think!

 

[bob1182006]

no what PowerIso did was this I think:

\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.

 

[camilus]

I don't 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

 

[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:

\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.

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

\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

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:

\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.

slick ;)

 

[camilus]

\int \limits_0^1 \frac{dx}{1+e^x} \approx 0.379885493

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

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. \frac{d}{du}\ln\left(u^{2}-u\right) \neq \frac{1}{u^{2}-u}.

 

[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?