Question

How do you integrate `int (6-15x)/( (x-1) (x+2) (x-4))` using partial fractions?

Answer 1

`color(red)(int (6-15x)/((x-1)(x+2)(x-4)) dx=)`

`color(red)(ln(x-1)+2ln(x+2)-3ln(x-4)+C_0)`

Explanation:

from the given integral, set up the equation using the variables A, B, and C

`int (6-15x)/((x-1)(x+2)(x-4)) dx=int(A/(x-1)+B/(x+2)+C/(x-4))dx`

`(6-15x)/((x-1)(x+2)(x-4))=A/(x-1)+B/(x+2)+C/(x-4)`

`(6-15x)/((x-1)(x+2)(x-4))=(A(x^2-2x-8)+B(x^2-5x+4)+C(x^2+x-2))/((x-1)(x+2)(x-4))`

`(6-15x)/((x-1)(x+2)(x-4))`
`=((A+B+C)x^2+(-2A-5B+C)x^1+(-8A+4B-2C)x^0)/((x-1)(x+2)(x-4))`

The equations will be

`A+B+C=0`
`-2A-5B+C=-15`
`-8A+4B-2C=6`

Simultaneous solution results to:

`A=1` and `B=2` and `C=-3`

final answer

`color(red)(int (6-15x)/((x-1)(x+2)(x-4)) dx=)`

`color(red)(ln(x-1)+2ln(x+2)-3ln(x-4)+C_0)`

God bless....I hope the explanation is useful.