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प्रश्न
Evaluate:
`int x/((x - 1)^2(x + 2)) dx`
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उत्तर
Let I = `int x/((x - 1)^2(x + 2)) dx`
Consider `x/((x - 1)^2(x + 2)) = A/(x - 1) + B/(x - 1)^2 + C/(x + 2)`
= `(A(x - 1)(x + 2) + B(x + 2) + C(x - 1))/((x - 1)^2(x + 2))`
∴ x = A (x − 1) (x + 2) + B (x + 2) + C (x − 1)2 ....(i)
Putting x = 1 in (i), we get
1 = A (0) (3) + B (3) + C (0)2
∴ 1 = 3B
∴ B = `1/3`
Putting x = −2 in (i), we get
−2 = A(−3) (0) + B (0) + C (9)
∴ −2 = 9C
∴ C = `-2/9`
Putting x = −1 in (i), we get
−1 = A(−2) (1) + B (1) + C (4)
∴ −1 = `-2A + 1/3 - 8/9 `
∴ −1 = `- 2A - 5/9`
∴ 2A = `- 5/9 + 1`
2A `= 4/9`
∴ A = `2/9`
∴ `x/((x - 1)^2(x + 2)) = ((2/9))/(x - 1) + ((1/3))/(x - 1)^2 + ((- 2/9))/(x + 2)`
∴ I = `int [((2/9))/(x - 1) + ((1/3))/(x - 1)^2 + ((- 2/9))/(x + 2)]` dx
= `2/9 int 1/(x - 1) dx + 1/3int (x - 1)^-2 dx - 2/9 int 1/(x + 2) dx`
= `2/9 log |x - 1| + 1/3 * (x - 1)^-1/((-1)) - 2/9 log |x + 2|` + c
= `2/9 log |x - 1| - 2/9 log |x + 2| - 1/3 xx 1/(x - 1) + c`
∴ I = `2/9 log |(x - 1)/(x + 2)| - 1/(3(x - 1))` + c
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