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.
 
54
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HallsofIvy

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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.
 
19
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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. [tex]\int{\frac{1+e^x-e^x}{1+e^x}dx}[/tex] [lets add 1 to our expression]

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

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

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

5. [tex]\int{1dx}-\int{\frac{du}{u}}[/tex]

6. [tex]x-ln|{u}|+C[/tex]

7. [tex]x-ln|{1+e^x}|+C[/tex] [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
 
534
1
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 [tex]u = e^{-x} + 1[/tex]:
[tex]
\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.
[/tex]

Note that
[tex]-\ln(e^{-x} + 1) = x - \ln(1 + e^x) = \ln \left( \frac{e^x}{1 + e^x} \right).[/tex]


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