How do you write the partial fraction decomposition of the rational expression # (y^2 + 1) / (y^3 - 1)#?

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Answer 1 · 2017-02-28T10:37:53

The answer is #=(2/3)/(y-1)+(1/3y-1/3)/(y^2+y+1)#

Let's factorise the denominator

#y^3-1=(y-1)(y^2+y+1)#

Therefore,

#(y^2+1)/(y^3-1)=(y^2+1)/((y-1)(y^2+y+1))#

Let's perform the decomposition into partial fractions

#(y^2+1)/((y-1)(y^2+y+1))=A/(y-1)+(By+C)/(y^2+y+1)#

#=(A(y^2+y+1)+(By+C)(y-1))/((y-1)(y^2+y+1))#

The denominators are the same, we can compare the numerators

#y^2+1=A(y^2+y+1)+(By+C)(y-1)#

Let #y=1#, #=>#, #2=3A#, #=>#, #A=2/3#

Coefficients of #y^2#

#1=A+B#, #=>#, #B=1-A=1-2/3=1/3#

Also,

#1=A-C#, #=>#, #C=A-1=2/3-1=-1/3#

Therefore,

#(y^2+1)/((y-1)(y^2+y+1))=(2/3)/(y-1)+(1/3y-1/3)/(y^2+y+1)#