Express 4#x^2#-x+3/#x^3#-1 in partial fractions. ?

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Answer 1 · 2018-07-03T17:00:06

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

The denominator is

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

Perform the decomposition into partial fractions

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

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

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

The denominators are the same, compare the numerators

#4x^2-x+3=A(x^2+x+1)+(Bx+C)(x-1)#

Let #x=1#, #=>#, #6=3A#, #=>#, #A=2#

Coefficients of #x^2#

#4=A+B#, #=># , #B=4-A=4-2=2#

Let #x=0#, #=>#, #3=A-C#,#=>#, #C=A-3=2-3=-1#

Finally,

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