How do you express # (7x-2)/((x-3)^2(x+1))# in partial fractions?

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Answer 1 · 2017-02-11T07:39:53

The answer is #=(19/4)/(x-3)^2+(9/16)/(x-3)+(-9/16)/(x+1)#

Let's perform the decomposition into partial fractions

#(7x-2)/((x-3)^2(x+1))=A/(x-3)^2+B/(x-3)+C/(x+1)#

#=(A(x+1)+B((x-3)(x+1))+C(x-3)^2)/((x-3)^2(x+1))#

The denominators are the same, we compare the numerators

#7x-2=A(x+1)+B((x-3)(x+1))+C(x-3)^2#

Let #x=3#, #=>#, #19=4A#, #=>#, #A=19/4#

Let #x=-1#, #=>#, #-9=16C#, #=>#, #C=-9/16#

Coefficients of #x^2#

#0=B+C#, #=>#, #B=-C=9/16#

Therefore,

#(7x-2)/((x-3)^2(x+1))=(19/4)/(x-3)^2+(9/16)/(x-3)+(-9/16)/(x+1)#