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Integrate the rational function: 1x2-9

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Question

Integrate the rational function:

`1/(x^2 - 9)`

Sum
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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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Chapter 7: Integrals - Exercise 7.5 [Page 322]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.5 | Q 2 | Page 322

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