Evaluate: `int(2x-1)/((x-1)(x+2)(x-3))dx`

Answer

N/a

Explanation

Let `=int((2x-1))/((x-1)(x+2)(x-3))dx`
Now, `(2x-1)/((x-1)(x+2)(x+3))= (A)/((x-1))+(B)/((x-2))+(C)/((x-3))`
`rArr 2x - 1 = A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)`
Put `x = 3`, then
`6-1 = C(3-1)(3+2)`
`rArr 2 = 10 C rArr C = 1/2`
Again but x = 1, then
`5 - 1 = A(1+2) (1-3)`
`rArr 1 = - 6A rArr = - 1/6`
Now, put `x = - 2`, then
`-4 - 1 = B(-2-1)(-2-3)`
`rArr -5 = 15 B rArr B = -1/3`
`:. I = -1/6 int (1)/(x-1) dx - 1/3 int (1)/(x+2)dx + 1/2 int (1)/(x-3) dx`
`= - 1/6 |(x-1)|-1/3 "log"|(x+2)|+ 1/2 "log" |(x-3) | + C`
`= - log |(x-1)|^(1//6) - log |(x+2)|^(1//3) + log |(x-3)|^(1//2) + C`
`= log|(sqrt(x-3))/((x-1)^(1//6)(x+2)^(1//3))|+C`