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Integrate the following w.r.t.x: x2(x-1)(3x-1)(3x-2) - Mathematics and Statistics

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प्रश्न

Integrate the following w.r.t.x:

`x^2/((x - 1)(3x - 1)(3x - 2)`

बेरीज

उत्तर

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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पाठ 3: Indefinite Integration - Miscellaneous Exercise 3 [पृष्ठ १५०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 3.14 | पृष्ठ १५०

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