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∫x2+x-1x2+x-6 dx

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Question

`int (x^2 + x -1)/(x^2 + x - 6)  "d"x`

Sum
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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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Chapter 2.3: Indefinite Integration - Short Answers II

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