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प्रश्न
`int ("d"x)/(x^3 - 1)`
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उत्तर
Let I = `int ("d"x)/(x^3 - 1)`
= `int 1/((x - 1)(x^2 + x + 1)) "d"x`
Let `1/((x - 1)(x^2 + x + 1))`
= `"A"/(x - 1) + ("B"x + "C")/(x^2 + x + 1)`
∴ 1 = A(x2 + x + 1) + (Bx + C)(x – 1) .......(i)
Putting x = 1 in (i), we get
1 = A(12 + 1 + 1)
∴ 1 = 3A
∴ A = `1/3`
Putting x = 0 in (i), we get
1 = A(0 + 0 + 1) + (0 + C)(0 – 1)
∴ 1 = A – C
∴ 1 = `1/3 - "C"`
∴ C = `- 2/3`
Putting x = 2 in (i), we get
1 = A(22 + 2 + 1) + (2B + C)(2 – 1)
∴ 1 = 7A + 2B + C
∴ 1 = `7/3 + 2"B" - 2/3`
∴ 1 = `5/3 + 2"B"`
∴ `(-2)/(3)` = 2B
∴ B = `-1/3`
∴ I = `int ((1/3)/(x - 1) + (-1/3x - 2/3)/(x^2 + x + 1)) "d"x`
= `1/3 int(1/(x - 1) - (x + 2)/(x^2 + x + 1)) "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/3 int (x + 2)/(x^2 + x + 1) "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/3*1/2 int (2x + 4)/(x^2 + x + 1) "d"x`
= `1/3 int 1/(x 1) "d"x - 1/6 int ((2x + 1) + 3)/(x^2 + x + 1)* "d"x`
= `1/3 int 1/(x - 1) "d"x - 1/6 int (2x + 1)/(x^2 + x + 1) "d"x - 1/2 int ("d"x)/(x^2 + x + 1)`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2 int ("d"x)/(x^2 + x + 1/4 - 1/4 + 1)` ......`[∵ int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + "c"]`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2 int ("d"x)/((x + 1/2)^2 + (sqrt(3)/2)^2`
= `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/2* 1/(sqrt(3)/2) tan^-1 ((x + 1/2)/(sqrt(3)/2)) + "c"`
∴ I = `1/3 log|x - 1| - 1/6 log|x^2 + x + 1| - 1/sqrt(3) tan^-1 ((2x + 1)/sqrt(3)) + "c"`
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