Question

How do you integrate `int (3x-4)/((x-1)(x+3)(x-6))` using partial fractions?

Answer 1

Partial fraction decomposition is:
`A/(x-1)+B/(x+3)+C/(x-6)`
Cover up rule or simultaneously solving by equating coefficients yields `A=1/20`, `B=-13/36`and `C=14/45`

`int(3x-4)/((x-1)(x+3)(x-6))dx`
`=1/20int1/(x-1)dx-13/36int1/(x+3)dx+14/45int(1/x-6)dx`
`=1/20ln|x-1|-13/36ln|x+3|+14/45ln|x-6|+c`