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Question
Integrate the rational function:
`1/(x^2 - 9)`
Solution
Let `1/(x^2 - 9) = 1/((x - 3)(x + 3))`
`= A/(x - 3) + B/(x + 3)`
⇒ 1 ≡ A(x + 3) + B(x - 3)
Put x = 3
1 = A (3 + 3)
⇒ A `= 1/6`
again, put x = -3
1 = B(3 - 3)
⇒ B `= -1/6`
`therefore 1/(x^2 - 9) = 1/6 [1/(x - 3) - 1/(x + 3)]`
`=> int 1/(x^2 - 9) = 1/6 int (1/(x - 3) - 1/(x + 3))` dx
`= 1/6 [log abs (x - 3) - log abs (x + 3)] + C`
`= 1/6 log abs ((x - 3)/(x + 3)) + C`
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