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प्रश्न
Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx
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उत्तर
Let I = `int ("x"^2 + "x" − 1)/("x"^2 + "x" − 6)` dx
`= int(("x"^2 + "x" − 6) + 5)/("x"^2 + "x" − 6)` dx
`= int [("x"^2 + "x" − 6)/("x"^2 + "x" − 6) + 5/("x"^2 + "x" − 6)]` dx
`= int [1 + 5/("x"^2 + "x" − 6)]` dx
`int [1 + 5/(("x + 3")("x − 2"))]` dx
Let `5/(("x + 3")("x − 2")) = "A"/"x + 3" + "B"/"x − 2"`
∴ 5 = A(x − 2) + B(x + 3) ....(i)
Putting x = 2 in (i), we get
5 = A (0) + B (5)
∴ 5 = 5B
∴ B = 1
Putting x = − 3 in (i), we get
5 = A(− 5) + B (0)
∴ 5 = − 5A
∴ A = − 1
∴ `5/(("x + 3")("x - 2")) = (-1)/"x + 3" + 1/"x − 2"`
∴ I = `int [1 + (-1)/"x + 3" + 1/"x − 2"]` dx
`= int "dx" - int 1/"x + 3" "dx" + int1/"x − 2"` dx
∴ I = x − log |x + 3| + log |x − 2| + c
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