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Integrate the following w.r.t. x : 12x+36x2+13x-63

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प्रश्न

Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`

योग
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उत्तर

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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.4 [पृष्ठ १४४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.4 | Q 1.03 | पृष्ठ १४४

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