Question

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

Answer 1

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

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 = `-2"A" + 1/3 - 8/9` 

∴ – 1 = `-2"A" - 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)] "d"x`

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

= `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"`