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e^x/(1+e^x)

Integral of e^x/(1+e^x) dx

Limits of integration:

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The graph:

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Piecewise:

The solution

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01exex+1dx\int\limits_{0}^{1} \frac{e^{x}}{e^{x} + 1}\, dx
Integral(E^x/(1 + E^x), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=exu = e^{x}.

      Then let du=exdxdu = e^{x} dx and substitute dudu:

      1u+1du\int \frac{1}{u + 1}\, du

      1. Let u=u+1u = u + 1.

        Then let du=dudu = du and substitute dudu:

        1udu\int \frac{1}{u}\, du

        1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

        Now substitute uu back in:

        log(u+1)\log{\left(u + 1 \right)}

      Now substitute uu back in:

      log(ex+1)\log{\left(e^{x} + 1 \right)}

    Method #2

    1. Let u=ex+1u = e^{x} + 1.

      Then let du=exdxdu = e^{x} dx and substitute dudu:

      1udu\int \frac{1}{u}\, du

      1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

      Now substitute uu back in:

      log(ex+1)\log{\left(e^{x} + 1 \right)}

  2. Now simplify:

    log(ex+1)\log{\left(e^{x} + 1 \right)}

  3. Add the constant of integration:

    log(ex+1)+constant\log{\left(e^{x} + 1 \right)}+ \mathrm{constant}


The answer is:

log(ex+1)+constant\log{\left(e^{x} + 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                           
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 |   E                /     x\
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exex+1dx=C+log(ex+1)\int \frac{e^{x}}{e^{x} + 1}\, dx = C + \log{\left(e^{x} + 1 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.02.0
The answer [src]
-log(2) + log(1 + E)
log(2)+log(1+e)- \log{\left(2 \right)} + \log{\left(1 + e \right)}
=
=
-log(2) + log(1 + E)
log(2)+log(1+e)- \log{\left(2 \right)} + \log{\left(1 + e \right)}
-log(2) + log(1 + E)
Numerical answer [src]
0.620114506958278
0.620114506958278
The graph
Integral of e^x/(1+e^x) dx

    Use the examples entering the upper and lower limits of integration.