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

Explanation

` int (x)/((x-1)(x-2)(x-3)) dx= I " (say) "`
`" Let " (x)/((x-1)(x-2)(x-3))`
`=(A)/((x-1))+(B)/((x-2))+(c)/((x-3))`
` rArr (x)/((x-1)(x-2)(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 " then " 1=A (-1)(-2)+0+0 rArr A=(1)/(2)`
` x= 2 " then " 2=0+B(-1)(1)+0 rArr B=-(1)/(2)`
`x=3 " then " 3=0 +0 +C(2)(1) rArr c=(3)/(2)`
` " Now " 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`
`-2 int (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)`