Solved Problems

Problem 2-8: Integration

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Yaz

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Find \displaystyle\int \! \frac{e^x}{1+e^x}\, dx.

Solution

Let u=1+e^x, therefore du=e^x dx \to dx=\frac{1}{u-1}du. Hence

\displaystyle\int \! \frac{e^x}{1+e^x}\, dx = \displaystyle\int \! \frac{u-1}{u} \frac{1}{u-1}\,du = \displaystyle\int \! \frac{1}{u} \,du=\ln (u)=\ln(1+e^x)
Problem 2-8

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    Comments

    3 responses to “Problem 2-8: Integration”

    1. jose flaminio tellez abril Avatar
      jose flaminio tellez abril

      aprender a integrar


      _____________
      learned to integrate

      Reply
    2. yohannes berhanu Avatar
      yohannes berhanu

      i really appreciate if u can send me newsletter every week. i am a pre college student , and i am glad to have this kind of website that would enable me and other fellas to succed on their academic activities.

      Reply
      1. Dr. Yaz Z. Li Avatar
        Dr. Yaz Z. Li

        Dear Yohannes,

        I am thinking of starting a newsletter. All registered user would be receiving it.

        Reply

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