Sum
Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx
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Solution
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 + 2) ....(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)
∴ 6 = - 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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