How do I integrate (x-1)/(x^2-1)?

Answer 1

use x^2-1=(x-1)(x+1)
thus
(x-1)/(x^2-1)=1/(x+1)=log(x+1)

Answer 2

1) First simplify, like others have mentioned. You get 1/(1+x). This step does carry a price, because we will have to check out what happens at x=1. The original function does not have a well-defined value at 1 or -1. More later.

2) Note that the derivative of f(g(x) is f'(g(x)*g'(x)dx. So if you can redefine the function in these terms, you are good to go.

3) Set g(x) = 1 + x. Set f'(x) = 1/x dx. Note that g'(x) = 1 dx and f(x) = ln(x)

4) Using our guessed-at functions, we see that f'(g(x)) = 1/(1+x) and g'(x) = 1, so that f(g(x) is f'(g(x)*g'(x)dx = 1/(1+x) dx

5) This is indeed the function we are trying to integrate, so we can accept the consequences and see what f(g(x)) is. f(g(x) = ln( 1+x ) + C. (C is any constant number. Because a constant number derives down to 0, integrating may leave us with just about any constant number, and we note that with C)

6) Checking this, we see that the derivative of ln( 1+x ) + C is indeed 1/(1+x) dx. Great!

7) Um, but now it gets a bit messier. ln(y) is undefined for y<=0, so we need to be careful here. Looking back at our function, we see that x <= -1 will not work with our found function. Oh well.

8) But what about that pesky x=1 and x=-1 problem from step 1? Well, x=-1 just petered out in step 7. We already know that -1 is not going to go well for our function. What about x=1? Looking back at our original function, x=1 would cause us to divide by 0, and that dog won't bark (Futurama quote :) ). Fortunately, we can quickly see that the function approaches 0.5 at x=1 from both sides, so it turns out to all be well.

9) To sum it up: the integral is ln(1+x) + C for x > -1 and undefined for x <= -1

Answer 3

1) Given
U(x) = (x-1)/(x^2-1) = (x-1)/[(x-1)(x+1)] = 1/(x+1)

2) Let G(x) be the anti derivative of U(x)
G(x) = ln(1+x)

3) Finally, the integral of U(x) is G(x) + C

Answer 4

If you simplify it, you get a much easier fraction.
(x^2-1) is (x-1)(x+1) It is not the same as (x-1)^2 or (x-1)(x-1)

It is 1/(x+1)
Now that would be easier to simplify, you might want to use substitution.

Answer 5

x-1/ (x-1)(x-1)= 1/x-1

Answer 6

I = ∫(x - 1) / [(x - 1).(x + 1)] dx
I = ∫1 / (x + 1) dx
I = log(x + 1) + c
I = k log (x + 1)

Answer 7

I = ∫(x - 1) / [(x - 1).(x + 1)] dx
I = ∫1 / (x + 1) dx
I = log(x + 1) + c
I = k log (x + 1)