Misc integral

  • Thread starter nameVoid
  • Start date
  • #1
241
0
[tex]
\int \frac{1+e^x}{1-e^x}dx
[/tex]
[tex]
\int \frac{dx}{1-e^x} +\int \frac{e^x}{1-e^x}dx
[/tex]
[tex]
u=e^x
[/tex]
[tex]
lnu=x
[/tex]
[tex]
\frac{du}{u}=dx
[/tex]
[tex]
\int \frac{du}{u(1-u)}+\int \frac{du}{1-u}
[/tex]
[tex]
\int \frac{A}{u}+\frac{B}{1-u}du -ln|1-u|+C
[/tex]
[tex]
ln|e^x|+ln|1-e^x|-ln|1-e^x|+C
[/tex]
[tex]
x+C
[/tex]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
34,886
6,626
[tex]
\int \frac{1+e^x}{1-e^x}dx
[/tex]
[tex]
\int \frac{dx}{1-e^x} +\int \frac{e^x}{1-e^x}dx
[/tex]
[tex]
u=e^x
[/tex]
[tex]
lnu=x
[/tex]
[tex]
\frac{du}{u}=dx
[/tex]
[tex]
\int \frac{du}{u(1-u)}+\int \frac{du}{1-u}
[/tex]
[tex]
\int \frac{A}{u}+\frac{B}{1-u}du -ln|1-u|+C
[/tex]
[tex]
ln|e^x|+ln|1-e^x|-ln|1-e^x|+C
[/tex]
[tex]
x+C
[/tex]
Your antiderivative is obviously incorrect, since d/dx(x + C) = 1. Your antiderivative would have been correct if its derivative was (1 + e^x)/(1 - e^x).

It might be easier not to split into two integrals, but using the same substitution. If you do that, you'll get something you can use partial decomposition on.
 
  • #3
241
0
[tex]
x-2ln|1-e^x|+C
[/tex]
 

Related Threads on Misc integral

  • Last Post
Replies
2
Views
737
  • Last Post
Replies
2
Views
862
  • Last Post
Replies
2
Views
919
  • Last Post
Replies
3
Views
876
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
855
  • Last Post
Replies
1
Views
864
  • Last Post
Replies
12
Views
763
Top