How do you express #(x+1 )/[ x^2(x-2)]# in partial fractions?

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Answer 1 · 2017-01-31T19:05:14

The answer is #=(-1/2)/x^2+(-3/4)/x+(3/4)/(x-2)#

Let's perform the decomposition into partial fractions

#(x+1)/(x^2(x-2))=A/x^2+B/x+C/(x-2)#

#=(A(x-2)+Bx(x-2)+Cx^2)/(x^2(x-2))#

As the denominators are the same, we can equalize the nimerators

#x+1=A(x-2)+Bx(x-2)+Cx^2#

Let #x=0#, #=>#, #1=-2A#, #=>#, #A=-1/2#

Let #x=2#, #=>#, #3=4C#, #=>#, #C=3/4#

Coefficients of #x^2#, #=>#, #0=B+C#

#C=-B=-3/4#

Therefore,

#(x+1)/(x^2(x-2))=(-1/2)/x^2+(-3/4)/x+(3/4)/(x-2)#