# What is domain of arctg 1/(x^2-9)?

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You need to remember that the domain of the arctangent function comprises all real numbers, hence, you need to consider the following condition for the principal argument `1/(x^2 - 9)` such that:

`1/(x^2 - 9) in R => x^2 - 9 != 0 => x^2 != 9 => x_(1,2) != +-3`

**Hence, evaluating the domain of the given function yields `x in R - {-3,3}.` **

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