`(x)/((x-1)(x-2)(x-3))`

Explanation

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