mash
how do you find intergal of 1/(y^2 -1)
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RadEn
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partial fraction
1/(y^2-1) = A/(y+1) + B/(y-1)
RadEn
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u have to find the values of A and B, first
completeidiot
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or trig substitution
completeidiot
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no sorry that would be a terrible idea
RadEn
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got it ?
RadEn
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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
RadEn
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use integral by u-sub
RadEn
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for int (-1/2)/(y+1)
let u = y+1
du = dy
it can be :
int (-1/2)/u du = -1/2 ln(u)
RadEn
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subtitute back that u = y+1
now, we have
int (-1/2)/u du = -1/2 ln(u) = -1/2 ln(y+1)
RadEn
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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)
RadEn
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finally, we have the answer :
-1/2 ln(y+1) + 1/2 ln(y-1) + c
RadEn
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does that make sense ?
mash
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perfect sense ty
RadEn
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you're welcome