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## mash 2 years ago how do you find intergal of 1/(y^2 -1)

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1. RadEn

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

2. RadEn

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

3. completeidiot

or trig substitution

4. completeidiot

no sorry that would be a terrible idea

5. RadEn

got it ?

6. RadEn

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

7. RadEn

use integral by u-sub

8. RadEn

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

9. RadEn

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

10. RadEn

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)

11. RadEn

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

12. RadEn

does that make sense ?

13. mash

perfect sense ty

14. RadEn

you're welcome

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