How do you integrate #int (x^3-3x^2-9)/(x^3-3x^2) dx# using partial fractions?

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Answer 1 · 2017-03-01T12:22:36

The answer is #=x-3/x+ln(|x|)-ln(|x-3|)+C#

We factorise the denominator

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

We start by performing a polynomial long division

#color(white)(aaaa)##x^3-3x^2##color(white)(aaaaaa)##-9##color(white)(aaaa)##|##x^3-3x^2#

#color(white)(aaaa)##x^3-3x^2##color(white)(aaaaaa)####color(white)(aaaaaaa)##|##1#

#color(white)(aaaaaa)##0-0##color(white)(aaaaaa)##-9##color(white)(aaaa)#

Therefore,

#(x^3-3x^2)/(x^3-3x^2)=1-9/(x^2(x-3))#

Now, we perform the partial fraction decomposition

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

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

The denominators are the same, we compare the numerators

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

Let #x=0#,#=>#,#9=-3A#, #=>#, #A=-3#

Let #x=3#, #=>#, #9=9C#, #=>#,#C=1#

Coefficients of #x^2#

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

Therefore,

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

So,

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

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

Then,

#int((x^3-3x^2)dx)/(x^3-3x^2)=int1dx+3intdx/(x^2)+intdx/(x)-intdx/(x-3)#

#=x-3/x+ln(|x|)-ln(|x-3|)+C#