Question

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

Answer 1

`-1/20ln|x-1|+13/4ln|x+3|-16/5ln|x+4|+C.`

Explanation:

Put `I=int(3x-4)/{(x-1)(x+3)(x+4)}dx.,` and, notice that the `Dr.` of the rational integrand is made of non-repeated linear factors. In such case, we have to find, `A,B,D in RR` such that,
`(3x-4)/{(x-1)(x+3)(x+4)}=A/(x-1)+B/(x+3)+D/(x+4)............(i)`
`={A(x+3)(x+4)+B(x-1)(x+4)+D(x-1)(x+3)}/{(x-1)(x+3)(x+4)}.`

As the `Drs.` of both sides are same, so must be the `Nrs.` Hence,

`A(x+3)(x+4)+B(x-1)(x+4)+D(x-1)(x+3)=3x-4...........(ii)`

Although `(ii)` holds good `AAx in RR`, but, we chose `x=1, x=-3, and, x=-4` [the reason behind this will soon be clear!] to get,
`A(4)(5)+B*0+D*0=3-4 rArr A=-1/20.`
`A*0+B(-4)(1)+D*0=-9-4 rArr B=13/4.`
`A*0+B*+D(-5)(-1)=-12-4 rArr D=-16/5.` Therefore, by `(i)`
`I=int(-1/20)/(x-1)dx+int(13/4)/(x+3)dx+int(-16/5)/(x+4)dx`
`I=-1/20ln|x-1|+13/4ln|x+3|-16/5ln|x+4|+C.`

Enjoy Maths.!