Problem 2-8: Integration


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

3 thoughts on “Problem 2-8: Integration

  1. 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.

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>