Question

Evaluate:

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

Answer 1

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

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

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

2A `= 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