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प्रश्न
`int (6x^3 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
उत्तर
Let I = `int (6x^2 + 5x^2 - 7)/(3x^2 - 2x - 1) "d"x`
2x + 3
`3x^2 - 2x - 1")"overline(6x^3 + 5x^2 + 0x - 7`
6x3 − 4x2 − 2x
(−) (+) (+)
9x2 + 2x − 7
9x2 − 6x − 3
(−) (+) (+)
8x − 4
∴ I = `int (2x + 3 + (8x - 4)/(3x^2 - 2x - 1)) "d"x`
3x2 – 2x – 1 = 3x2 – 3x + x – 1
= 3x(x – 1) + 1(x – 1)
= (x – 1)(3 x + 1)
∴ I = `int[2x + 3 + (8x - 4)/((x - 1)(3x + 1))] "d"x`
Let `(8x - 4)/((x - 1)(3x + 1)) = "A"/(x - 1) + "B"/(3x + 1)`
∴ 8x – 4 = A(3x + 1) + B(x – 1) ........(i)
Putting x = 1 in (i), we get
4 = 4A
∴ A = 1
Putting x = `(-1)/3` in (i), we get
`8(-1/3) - 4 = "B"(-1/3 - 1)`
∴ `(-20)/3 = -4/3 "B"`
∴ B = 5
∴ `(8x - 4)/((x - 1)(3x + 1)) = 1/(x - 1) + 5/(3x + 1)`
∴ I = `int (2x + 3 + 1/(x - 1) + 5/(3x + 1)) "d"x`
= `2 int x "d"x + 3 int "d"x + int 1/(x - 1) "d"x + 5/3 int 3/(3x + 1) "d"x`
= `2(x^2/2) + 3x + log|x + 1| + (5log|3x + 1|)/3 + "c"`
∴ I = `x^2 + 3x + log|x - 1| + 5/3 log|3x + 1| + "c"`
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