Question

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

Answer 1

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`