Question

how do you find intergal of 1/(y^2 -1)

Answer 1

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

Answer 2

u have to find the values of A and B, first

Answer 3

or trig substitution

Answer 4

no sorry that would be a terrible idea

Answer 5

got it ?

Answer 6

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

A+B = 0
B-A = 1
by using elimination or subtitution, we get B=1/2 and A = -1/2

see ur integral becomes
int (-1/2)/(y+1) dy + int(1/2)/(y-1) dy

Answer 7

use integral by u-sub

Answer 8

for int (-1/2)/(y+1)
let u = y+1
du = dy
it can be :
int (-1/2)/u du = -1/2 ln(u)

Answer 9

subtitute back that u = y+1
now, we have
int (-1/2)/u du = -1/2 ln(u) = -1/2 ln(y+1)

Answer 10

by same idea to solving int(1/2)/(y-1) dy
u = y-1
du = du

so, we have int(1/2)/u = 1/2 ln(u) = 1/2 ln(y-1)

Answer 11

finally, we have the answer :
-1/2 ln(y+1) + 1/2 ln(y-1) + c

Answer 12

does that make sense ?

Answer 13

perfect sense ty

Answer 14

you're welcome