Question

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

Answer 1

`(x+5)/((x+3)(x-2)(x-5))=A/(x+5)+B/(x-2)+C/(x-5)`
`x^2(A+B+C)-x(7A+2B+6C)+10A-15B-6C=x+5`
`A+B+C=0; 7A +2B+C=1; 10A-15B-6C=5`
`A=1/5; B=-1/5; C=0`
`int((x+5)/((x+3)(x-2)(x-5)))=int1/(5(x+3))-int1/(5(x-2))=1/5(ln(x+3)-ln(x-2))=1/5ln((x+3)/(x-2))`

Answer 2

`=1/20ln(x+3)-7/15ln(x-2)+5/12ln(x-5)+C`

Explanation:

Let's do the partial fractions first
`(x+5)/((x+3)(x-2)(x-5))=A/(x+3)+B/(x-2)+C/(x-5)`
`=(A(x-2)(x-5)+B(x+3)(x-5)+C(x+3)(x-2))/((x+3)(x-2)(x-5))`
so have the following
`x+5=A(x-2)(x-5)+B(x+3)(x-5)+C(x+3)(x-2)`
let `x=2` so `7=0-15B+0` so `B=-7/15`
let `x=5` so `10=0+0+24C` => `C=5/12`
let `x=-3` so `2=40A+0+0` => `A=1/20`
finally `int((x+5)dx)/((x+3)(x-2)(x-5))=1/20intdx/(x+3)-7/15intdx/(x-2)+5/12intdx/(x-5)`
`=1/20ln(x+3)-7/15ln(x-2)+5/12ln(x-5)+C`