How to integrate 1/(x^2-1)

1 / ( x² - 1) = 1 / [ (x - 1)(x + 1) ]

1 / [ ( x - 1)(x + 1) ] = A / (x - 1) + B / (x + 1)
1 = A (x + 1) + B (x - 1)

1 = 2A
A = 1/2

1 = -2B
B = - 1/2

I = (1/2) ∫ dx / ( x - 1) - (1/2) ∫ dx / ( x + 1)

I = (1/2) log ( x - 1 ) - (1/2) log ( x + 1 ) + C

Use Partial fraction decomposition.

Did you go to class?
Remember that nonsense about A/(x+1) + B(x-1) = 1/((x+1)(x-1)) ??
where you let x = 1, giving A= 1/2
x = -1, B = -1/2

then you should also remember that int(1/x)dx = ln |x| + C
ans that ln x - ln y = ln x/y

Or you can keep getting answers done for you like the douche bag's post above.
Do your own work.
good day