Question

How do you integrate `(2x-82)/(x^2+2x-48) dx` using partial fractions?

Answer 1

7ln(x+8)-5ln(x-6)+C

Explanation:

`int(2x-82)dx/(x^2+2x-48)`
`(2x-82)/(x^2+2x-48)=(2x-82)/((x-6)(x+8))=A/(x-6)+B/(x+8) =(A(x+8) +B(x-6))/((x-6)(x+8))`
`2x-82=A(x+8) +B(x-6)`
`if x=6 ; -70=14A ; A=-5`
`if x=-8 ; 98=-14B , B=7`
`int(2x-82)dx/(x^2+2x-48)=int7dx/(x+8)-int5dx/(x-6`
`=7ln(x+8)-5ln(x-6)+C`