Evaluate this Integral?

Question

∫2dx/(x² -1)

Answer 1

use need to use partial fractions

2/(x^2-1)
=2/[(x+1)(x-1)]...........(1)
=a/(x+1)+b/(x-1)..........(2)
solve for a and b
multiply (1) and (2) by (x+1)(x-1)
2=a(x-1)+b(x+1)
let x=1
2=0+2b
b=1
let x=-1
2=-2a+0
a=-1
therfore 2/(x^2-1)=-1/(x+1)+1/(x-1)

∫2dx/(x² -1)
=∫-1/(x+1)+1/(x-1) dx
=-In(x+1)+In(x-1)+c
=In[(x-1)/(x+1)]+c

Answer 2

1/(x^2-1) = a/(x-1)+b/(x+1) partial fractions
1=a(x+1)+b(x-1) so a=1/2 and b =-1/2
Int = ln I (x-1)/(x+1)I