# Seemingly simple integration

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?

## The Attempt at a Solution

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Tom Mattson
Staff Emeritus
Gold Member
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.

Mark44
Mentor
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 $\int du/u?$
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
$$\int \frac{dx}{f(x)} = ln |f(x)| + C$$

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

## The Attempt at a Solution

HallsofIvy
Homework Helper
In any case, please try partial fractions and get back to us with the result.

$$\frac{1}{x^2- 1}= \frac{1}{(x-1)(x+1)}= \frac{A}{x-1}+ \frac{B}{x+1}$$

What are A and B?