how would one integrate 1/(x^2 - 1) ?

Question

∫1/(x^2 - 1) dx
= ∫(1/2)[1/(x-1) - 1/(x+1)]dx

how do you get this step so quickly?

The way i tried, involved using partial fractions:

∫1/(x^2 - 1) dx
=∫1/(x-1)(x+1) dx
=∫A/(x-1) +B(x+1) dx
=∫[ A(x+1) + B(x-1)]/[(x+1)(x-1)]dx
equating numerators:
1 = A(x+1) + B(x-1)
A = 1/2
B = -1/2
=∫1/2/(x-1) - 1/2(x+1) dx
=1/2∫1/(x-1) - 1/(x+1) dx
= 1/2ln[(x-1)/(x+1)] + c

Answer 1

∫1/(x^2 - 1) dx
= ∫(1/2)[1/(x-1) - 1/(x+1)]dx
= (1/2)ln|(x-1)/(x+1)| + c
---------
It is not necessary to do partial fraction in a standard way every time. For simple problems such as this one, you can do it mentally. You separate 1/(x^2-1) into two parts 1/(x+1) and 1/(x-1) first. Since you want to get the numerator "1", you must have 1/(x-1) - 1/(x+1) to cancel x in the numerator, and then multiply it by (1/2).
1/(x-1) - 1/(x+1) = 2/(x^2-1)
(1/2)2/(x^2-1) = 1/(x^2-1)

Answer 2

Remember d/dx ln(x) = 1/x

=> ∫ 1/(x^2 – 1) = ln(x^2 - 1)

Answer 3

Using partial fractions,

INT 1/(x^2-1) = INT 1/(x-1)(x+1)

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

,', A(x+1) + B(x-1) = 1

let x = 1

2A = 1
A = 1/2

let x = -1

-2B = 1
B = -1/2

.'. int 1/(x^2 - 1) = int (1/2(x-1) - 1/2(x+1))

= 1/2 int 1/x-1 - 1/x+1
= 1/2 [ ln(x-1) - ln(x+1)]
= 1/2 ln (x-1/x+1) +c

Answer 4

1 / (x ² - 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) ∫ 1 / (x - 1) dx - (1/2) ∫ 1 / 9x + 1) dx
I = (1/2) log (x - 1) - (1/2) log (x + 1) + C
Let C = log K
I = log K (x - 1)^(1/2) / (x + 1)^(1/2)
i = log K [ (x - 1) / (x + 1) ]^(1/2)