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

1 Answer
Oct 29, 2016

THe answer is #=x-2/x+lnx-ln(x-2)+C#

Explanation:

Let's start by rewriting the expression
#(x^3-2x^2-4)/(x^3-2x^2)=1-4/(x^3-2x^2)=1-4/((x^2)(x-2))#

So now we can apply the decomposition into partial fractions
#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))#
Solving for A,B and C

#1=A(x-2)+Bx(x-2)+Cx^2#
let #x=2##=>##1=4C##=>##C=1/4#
let #x=0##=>##1=-2A##=>##A=-1/2#
Coefficients of #x^2##=>##0=B+C##=>##B=-1/4#
so we have, #1/((x^2)(x-2))=-1/(2x^2)-1/(4x)+1/(4(x-2)#
So #1-4/((x^2)(x-2))=1-2/(x^2)-1/(x)+1/(x-2)#

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

#=x-2/x+lnx-ln(x-2)+C#