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Question
Integrate the following w.r.t.x:
`x^2/((x - 1)(3x - 1)(3x - 2)`
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Solution
Let I = `int x^2/((x - 1)(3x - 1)(3x - 2)).dx`
Let `x^2/((x - 1)(3x - 1)(3x - 2)) = "A"/(x - 1) + "B"/(3x -1) + "C"/(3x - 2)`
∴ x2 = A(3x -1)(3x - 2) + B(x – 1)(3x - 2) + C(x – 1)(3x -1)
Put x – 1 = 0, i.e. x = 1, we get
∴ x2 = A(2)(1) + B(0)(1) + C(0)(2)
∴ 2 = 4A
∴ A = `(1)/(2)`
Put x + 2 = 0, i.e. x = – 2, we get
2 + 2 = A(0)(1) + B(– 3)(1) + C(– 3)(0)
∴ 6 = – 3B
∴ B = – 2
Put x + 3 = 0, i.e. x = – 3we get
9 + 2 = A(– 1)(0) + B(– 4)(0) + C(– 4)(– 1)
∴ 11 = 4C
∴ C = `(11)/(4)`
∴ `(x^2 + 2)/((3x - 1)(x - 1)(3x - 2))`
= `((1/4))/(3x - 1) + (-2)/(x - 1) + ((11/4))/(3x - 2)`
∴ I = `int [((1/4))/(3x - 1) + (-2)/(x - 1) + ((11/4))/(3x - 2)].dx`
= `(1)/(18) int (1)/(3x - 1).dx - 2 int(1)/(x - 1).dx + (4)/(9) int (1)/(3x - 2).dx`
= `(1)/(18) log|3x - 1| + (1)/(2) log|x - 1| - (4)/(9) log|3x - 2| + c`.
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