Integrating 1/(x^2 - 1): Confusion with Online Calculator

[abite]

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?

 

[quantumdude]

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]

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?

 

[HallsofIvy]

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?