anonymous
  • anonymous
how do you find intergal of 1/(y^2 -1)
Mathematics
katieb
  • katieb
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RadEn
  • RadEn
partial fraction 1/(y^2-1) = A/(y+1) + B/(y-1)
RadEn
  • RadEn
u have to find the values of A and B, first
anonymous
  • anonymous
or trig substitution

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anonymous
  • anonymous
no sorry that would be a terrible idea
RadEn
  • RadEn
got it ?
RadEn
  • 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
RadEn
  • RadEn
use integral by u-sub
RadEn
  • 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)
RadEn
  • RadEn
subtitute back that u = y+1 now, we have int (-1/2)/u du = -1/2 ln(u) = -1/2 ln(y+1)
RadEn
  • 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)
RadEn
  • RadEn
finally, we have the answer : -1/2 ln(y+1) + 1/2 ln(y-1) + c
RadEn
  • RadEn
does that make sense ?
anonymous
  • anonymous
perfect sense ty
RadEn
  • RadEn
you're welcome

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