How do you integrate #int (x^3-x^2+1) / (x^4-x^3)# using partial fractions?

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Answer 1 · 2018-02-04T07:17:22

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

Perform the decomposition into partial fractions

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

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

#=(A(x-1)+B(x(x-1))+C(x^2(x-1))+D(x^3))/((x^4-x^3))#

The denominators are the same, compare the numerators

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

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

Coefficients of #x#

#0=A-B#, #=>#, #B=A=-1#

Coefficients of #x^2#

#-1=B-C#, #=>#, #C=B+1=-1+1=0#

Coefficients of #x^3#

#1=C+D#, #D=1-C=1-0=1#

Therefore,

#(x^3-x^2+1)/(x^4-x^3)=-1/(x^3)-1/(x^2)+0/(x)+1/(x-1)#

So,

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

#=1/(2x^2)+1/x+ln(|x-1|)+C#