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”

jose flaminio tellez abril September 30, 2010 at 3:18 am

aprender a integrar

_____________
learned to integrate

Reply ↓
yohannes berhanu January 28, 2011 at 6:18 pm

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 January 29, 2011 at 10:37 pm
  
  Dear Yohannes,
  
  I am thinking of starting a newsletter. All registered user would be receiving it.
  
  Reply ↓

Solved Problems in Calculus

A Source of Free Solved Problems in Calculus

Problem 2-8: Integration

3 thoughts on “Problem 2-8: Integration”

Leave a Reply Cancel reply