Evaluate : `int(x)/((x-1)^(2)(x+2))dx`.

Explanation

`" Let " (x)/((x-1)^(2)(x+2))`
`=(A)/(x-1)+(B)/(x-1)^(2)+(C)/(x+2) .........(1)`
` rArr x=A (x-1) (x+2) +B(x+2) +C(x-1)^(2).....(2)`
`x=-2 " then " -2=0 +0+C(9) rArr C=-(2)/(9)`
Equating coefficients of `x^(2)`,
`0=A +C rArr A=-C= (2)/(9)`
`:. int (x)/((x-1)^(2)(x+2))dx`
`=int{((2)/(9))/(x-1)+((1)/(3))/((x+1)^(2))+((-(2)/(9)))/(x+2)}dx`
[From equation (1)]
`= (2)/(9) int (1)/(x-1)dx +(1)/(3) int (1)/((x-1)^(2))dx`
`-(2)/(9)int(1)/(x+2)dx`
`=(2)/(9) log |x-1|+(1)/(3) ((x-1)^(-2+1))/((-2+1))`
`-(2)/(9) log |x+2|+C`
`=(2)/(9) log |(x-1)/(x+2)|-(1)/(3)((1)/(x-1))+C`