# Answer to Question #18237 in Calculus for johan

Question #18237

Solve the following integrals using partial fraction technique:

∫dy/(y+1)(y-1)

∫dy/(y+1)(y-1)

Expert's answer

1/((y-1)(y+1))=A/(y-1)+B(y+1)=(Ay+A+By-B)/(y-1)(y+1) =>A=1/2, B=-1/2

1/((y-1)(y+1))=1/2(y-1)-1/2(y+1)

Int(1/(y-1)(y+1))dy=Int(dy/2(y-1))-Int(dy/(y+1))=ln(y-1)/2+ln(y+1)/2+const=ln(sqrt((y-1)/(y+1)))+const

Here's another example from us about partial fractions method. take a look!

1/((y-1)(y+1))=1/2(y-1)-1/2(y+1)

Int(1/(y-1)(y+1))dy=Int(dy/2(y-1))-Int(dy/(y+1))=ln(y-1)/2+ln(y+1)/2+const=ln(sqrt((y-1)/(y+1)))+const

Here's another example from us about partial fractions method. take a look!

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