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Question
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
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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 [1 + (5)/(x^2 + x - 6)].dx`
= `int 1 dx + 5 int (1)/(x^2 + x - 6).dx`
Let `(1)/(x^2 + x - 6)`
= `(1)/((x + 3)(x - 2)`
= `"A"/(x + 3) + "B"/(x- 2)`
∴ 1 = A(x – 2) + B(x + 3)
Put x 3 = 0, i.e. x = –3, we get
1 = A(– 5) + B(0)
∴ A = `(-1)/(5)`
Put x – 2 = 0, i.e. x = 2, we get
1 = A(0) + B(5)
∴ B = `(1)/(5)`
∴ `(1)/(x^2 + x - 6) = ((-1/5))/(x + 3) + ((1/5))/(x - 2)`
∴ I = `int 1 dx + 5 int [((-1/5))/(x + 3) + ((1/5))/(x - 2)].dx`
= `int 1 dx - int (1)/(x + 3).dx + int (1)/(x - 2).dx`
= x – log|x + 3| + log|x – 2| + c
= `x + log|(x - 2)/(x + 3)| + c`.
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