# Evaluate: ∫x(x - 1)2(x + 2) dx - Mathematics and Statistics

Sum

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

#### Solution

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

Is there an error in this question or solution?
Chapter 5: Integration - Exercise 5.6 [Page 135]

Share