`(1)/(x^(2)-9)`

Explanation

`int(1)/(x^(2)-9)dx=int(1)/(x^(2)-3^(2))dx`
`" "=int(1)/((x+3)(x-3))dx=I" "` (माना )
माना `(1)/((x+3)(x-3))=(A)/((x+3))+(B)/((x-3))`
`rArr" "1=A(x-3)+B(x+3)`
`{:(x=3,"तो ",1=0+B(3+3),rArr.,B=(1)/(6)),(x=-3,"तो ",1=A(-3-3)+0,rArr.,A=-(1)/(6)):}`
`therefore" "I=int(1)/((x+3)(x-3))dx`
`=int((-1))/(6(x+3))dx+int(1)/(6(x-3))dx`
`=-(1)/(6)log|x+3|+(1)/(6)log|x-3|+C`
`=(1)/(6)log|(x-3)/(x+3)|+C`