# Integrating 1/(1+e^x)

Integrating 1/(1+e^x):

I am currently substituting 1+e^x = u,

S (u*e^x)^-1 du since du/dx = e^x

S (u*(u-1))^-1 du (S represents the integral symbol)

But i looked up the answer and got Ln (e^x(1+e^x)) +c

i need to understand how this is done, thanks in advance everyone.

HallsofIvy
Homework Helper
After you have integral (u(u-1))^(-1) du= integral 1/(u(u-1)) du

you can use "partial fractions", rewriting

1/(u(u-1))= A/u+ B/(u-1) for proper A and B. Then integrate

integral (A/u)du+ integral B/(u-1) du.

Guys this problem is a lot easier than it looks. From a quote from a PHD Student at Texas Tech Univ. Krishna Kaphle "Adding 0 or Multiplying By 1 is Free In Math".

What does this mean for this problem well lets rewrite our expression
1. $$\int{\frac{1+e^x-e^x}{1+e^x}dx}$$ [lets add 1 to our expression]

2. $$\int{\frac{[1+e^x]-e^x}{1+e^x}dx}$$ [now focus in on the numerator see whats in brackets matches the denominator]

3. $$\int{\frac{1+e^x}{1+e^x}dx}-\int{\frac{e^x}{1+e^x}dx}$$ [now lets split the fraction]

4. $$\int{1dx}-\int{\frac{e^x}{1+e^x}dx}$$ [now just use substiution method for right hand integral. $$u=1+e^x, du = e^xdx$$]

5. $$\int{1dx}-\int{\frac{du}{u}}$$

6. $$x-ln|{u}|+C$$

7. $$x-ln|{1+e^x}|+C$$ [substitue u back]

If you don't believe this or too lazy to do its derivative use this tool:
http://library.wolfram.com/webMathematica/Education/WalkD.jsp
and enter this expression: x-ln[1+e^x]

kiko

If the poster above can bump this thread, so can I. Besides, I think my solution is the nicest one between this thread and the thread referred to in post #2. :)

I make the substitution $$u = e^{-x} + 1$$:
$$\int \frac{1}{1 + e^x} \,dx = \int \frac{e^{-x}}{e^{-x} + 1} \,dx = \int -\frac{1}{u} \,du = -\ln \lvert u \rvert + C = -\ln(e^{-x} + 1) + C.$$

Note that
$$-\ln(e^{-x} + 1) = x - \ln(1 + e^x) = \ln \left( \frac{e^x}{1 + e^x} \right).$$

edit: Isn't this thread supposed to be locked, being archived?

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