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Question
`int (x^2 + x -1)/(x^2 + x - 6) "d"x`
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Solution
Let I = `int (x^2 + x -1)/(x^2 + x - 6) "d"x`
= `int (x^2 + x - 6 + 5)/(x^2 + x - 6) "d"x`
= `int [1 + (5/(x^2 + x - 6))] "d"x`
Let `5/(x^2 + x - 6) = 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 = B(5)
∴ B = 1
Putting x = −3 in (i), we get
5 = A(− 5)
∴ A = −1
∴ `5/((x + 3)(x - 2)) = (-1)/(x + 3) + 1/(x - 2)`
∴ I = `int[1 + (-1)/(x + 3) + 1/(x - 2)] "d"x`
= `int "d"x - int 1/(x + 3) "d"x + int 1/(x - 2) "d"x`
= x − log|x + 3| + log|x − 2| + c
∴ I = `x + log |(x - 2)/(x + 3)| + "c"`
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