Seemingly simple integration

  • Thread starter abite
  • Start date
  • #1
1
0
Homework Statement

I need to find the integral of 1/(x2 - 1) dx


The attempt at a solution

Double checking on an online integrator, it gave me an answer of

1/2 [log(x-1) - log(x+1)]

I would have expected

log(x2-1)/2


Does anyone know why it's the first one?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
8
Double checking on an online integrator,
That's fine, but have you tried it? This integral can easily be done either by partial fractions or trig substitution.
 
  • #3
34,172
5,787
Homework Statement

I need to find the integral of 1/(x2 - 1) dx


The attempt at a solution

Double checking on an online integrator, it gave me an answer of

1/2 [log(x-1) - log(x+1)]

I would have expected

log(x2-1)/2
Why would you have expected this? Were you thinking that your integral looked like [itex]\int du/u?[/itex]
If you were thinking along those lines, with u = x^2 - 1, you don't have anything remotely close to du to make this work.

It is NOT true that
[tex]\int \frac{dx}{f(x)} = ln |f(x)| + C[/tex]

If you weren't thinking that, never mind...
Does anyone know why it's the first one?

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
In any case, please try partial fractions and get back to us with the result.

[tex]\frac{1}{x^2- 1}= \frac{1}{(x-1)(x+1)}= \frac{A}{x-1}+ \frac{B}{x+1}[/tex]

What are A and B?
 

Related Threads on Seemingly simple integration

Replies
2
Views
1K
Replies
8
Views
2K
Replies
5
Views
775
Replies
6
Views
1K
Replies
9
Views
1K
Replies
2
Views
1K
Replies
8
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
3
Replies
52
Views
3K
Top