There are 2 ways to go here

1. Use partial fraction decomposition

1/(y^2-1) = 1/(y-1)(y+1) = A/(y-1) + B/(y+1)

1 = A(y+1) + B (y-1)

if y= 1 A =1/2 If y = -1 B= -1/2

1/(y^-1) = 1/2[1/(y-1) - 1/(y+1)]

you have 2 simple integrals

2. Trig sub

y =sec(t) dy = sec(t)tan(t) y^2 - 1 = tan(t)

You have int[ sec(t)/tan(t)] = int [ csc(t)] = -ln|csc(t) +cot(t)| +c

convert back to fn of y by drawing right triangle with t = arcsec(y)

so you're triangle has hypotneuse y adjacent side 1 opposite side

(y^2 -1)^(1/2)