Question

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

Answer 1

Let I = `int "x"/(("x - 1")^2("x + 2"))` dx

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

∴ x = A (x - 1) (x + 2) + B (x + 2) + C (x - 1)²  ....(i)

Putting x = 1 in (i), we get 

1 = A (0) (3) + B (3) + C (0)² 

∴ 1 = 3B

∴ B = `1/3`

Putting x = -2 in (i), we get

- 2 = A(- 3) (0) + B (0) + C (9)

∴ - 2 = 9C

∴ C = - `2/9`

Putting x = - 1 in (i), we get

- 1 = A(- 2) (1) + B (1) + C (4)

∴ - 1 = - 2A +`1/3 - 8/9 `

∴ - 1 = - 2A `- 5/9`

∴ 2A = `- 5/9 + 1 = 4/9`

∴ A = `2/9`

∴ `"x"/(("x - 1")^2("x + 2")) = (2/9)/"x - 1" + (1/3)/("x - 1")^2 + (- 2/9)/"x + 2"`

∴ I = `int [(2/9)/"x - 1" + (1/3)/("x - 1")^2 + (- 2/9)/"x + 2"]` dx

`= 2/9 int 1/"x - 1" "dx" + 1/3int ("x - 1")^-2 "dx" - 2/9 int 1/"x + 2" "dx"`

`= 2/9 log |"x - 1"| + 1/3 * ("x - 1")^-1/-1 - 2/9 log |"x + 2"|` + c

`= 2/9 log |"x - 1"| - 2/9 log |"x + 2"| - 1/3 xx 1/("x - 1") + "c"`

∴ I = `2/9 log |("x - 1")/("x + 2")| - 1/(3("x - 1"))` + c