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