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Question
Integrate the following w.r.t. x:
`(6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1)`
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Solution
Let I = `int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1).dx`
`3x^2 - 2x - 1")"overline(6x^3 + 5x^2 - 7)("2x + 3`
`6x^3 - 4x^2 - 2x`
– + +
`9x^2 + 2x - 7`
`9x^2 - 6x - 3`
– + +
8x – 4
∴ I = `int [(2x + 3) + (8x - 4)/(3x^2 - 2x - 1)] dx`
= `int 2x + 3 + int (8x - 4)/((x - 1)(3x + 1)) dx`
Let `(8x - 4)/((x - 1)(3x + 1)) = "A"/(x - 1) + "B"/(3x + 1)`
∴ 8x – 4 = A(3x + 1) + B(x – 1)
Put x – 1 = 0, i.e. x = 1, we get
8 – 4 = A(4) + B(0)
∴ A = 1
Put 3x + 1 = 0, i.e. x = `-(1)/(3)`, we get
`8(-1/3) - 4 = "A"(0) + "B"(-4/3)`
∴ `(-8 - 12)/(3) = -(4"B")/(3)`
∴ B = 5
∴ `(8x - 4)/((x - 1)(3x + 1)) = (1)/(x - 1) + (5)/(3x + 1)`
∴ `I = 2 int x dx + 3 int 1 dx + int [(1)/(x - 1) + (5)/(3x + 1)] dx`
= `2 (x^2/2) + 3x + int (1)/(x - 1) dx + 5 int (1)/(3x + 1) dx`
= `x^2 + 3x + log |x - 1| + 5/3 log |3x + 1| + c`.
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