Question

Evaluate the following:

`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`

Answer 1

Let I = `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`

Resolving into partial fraction, we put

`(2x - 1)/((x - 1)(x + 2)(x - 3)) = "A"/(x - 1) + "B"/(x + 2) + "C"/(x - 3)`

⇒ 2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)

Put x = 1

1 = A(3)(– 2)

⇒ A = `-1/6`

Put x = – 2

– 5 = B(– 3)(– 5)

⇒ B = `- 1/3`

Put x = 3

5 = C(2)(5)

⇒ C = `1/2`

∴ `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x = - 1/6 int 1/(x - 1) "d"x - 1/3 int 1/(x + 2) "d"x + 1/2 int 1/(x - 3) "d"x`

= `- 1/6 log |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)^(1/3) + "C"`

Hence, `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x = log[sqrt(x - 3)/((x - 1)^(1/6) (x + 2)^(1/3))] + "C"`