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Question
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
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Solution
Let I = `int (12x + 3)/(6x^2 + 13x - 63).dx`
Let `(12x + 3)/(6x^2 + 13x - 63)`
= `(12x + 3)/((2x + 9)(3x - 7)`
= `"A"/(2x + 9) + "B"/(3x - 7)`
∴ 12 + 3 = A(3x - 7) + B(2x + 9)
Put 2x + 9 = 0, i.e. x = `(-9)/(2)`, we get
`12((-9)/2) + 3 = "A"((-27)/2 - 7)+ "B"(0)`
∴ – 51 = `(-41)/(2)"A"`
∴ A = `(102)/(41)`
Put 3x – 7 = 0, i.x = `(7)/(3)`, we get
`12(7/3) + 3 = "A"(0) + "B"(14/3 + 9)`
∴ 31 = `(41)/(3)"B"`
∴ B = `(93)/(41)`
∴ `(12x + 3)/(6x^2 + 13x - 63)``(12x + 3)/(6x^2 + 13x - 63) = ((102/41))/(2x + 9) + ((93/41))/(3x - 7)`
∴ I = `int [((102/41))/(2x + 9) + ((93/41))/(3x - 7)].dx`
= `(102)/(41) int 1/(2x + 9).dx + 93/41 int 1/(3x - 7).dx`
= `(102)/(41).(log|2x + 9|)/(2) + 93/41.(log|3x - 7|)/(3) + c`
= `(51)/(41)log|2x + 9| + (31)/(41) log|3x - 7| + c`.
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