Question

How do you integrate `int (x^3-x^2-5x )/( x^2-3x+2)` using partial fractions?

Answer 1

The integral is `=x^2/2+2x+5ln(x-1)-6ln(x-2)+C`

Explanation:

As the degree of the numerator is greater than the denominator, let's do a long division
`x^3-x^2-5x` `color(white)(aaaaaa)` `∣` `x^2-3x+2`
`x^3-3x^2+2x` `color(white)(aaaaa)` `∣` `x+2`
`0+2x^2-7x`
`color(white)(aaaa)` `2x^2-6x+4`
`color(white)(aaaaaa)` `0-x-4`

We can factorise the denominator
`x^2-3x+2=(x-1)(x-2)`

So putting all together
`(x^3-x^2-5x)/(x^2-3x+2)=x+2+(-x-4)/(x^2-3x+2)`

Going to the decomposition into partial fractions
`(-x-4)/(x^2-3x+2)=(-x-4)/((x-1)(x-2))=A/(x-1)+B/(x-2)`
So `-x-4=A(x-2)+B(x-1)`
let `x=1` `=>` `-5=-A` so `A=5`
let `x=2` `=>` `-6=B` so `B=-6`
And we have
`(x^3-x^2-5x)/(x^2-3x+2)=x+2+5/(x-1)-6/(x-2)`
and the integral is
`int((x^3-x^2-5x)dx)/(x^2-3x+2)=int(x+2+5/(x-1)-6/(x-2))dx`
`=x^2/2+2x+5ln(x-1)-6ln(x-2)+C`