Question

How do you integrate `int sqrt(-x^2-6x-25)/xdx` using trigonometric substitution?

Answer 1

The function: `sqrt(-x^2-6x-25)/x` is not defined in `RR`

Explanation:

Note that:

`-x^2-6x-25 = -(x+3)^2 -16 < 0` for every `x`.

The function:

`sqrt(-x^2-6x-25)/x` is therefore not defined for any `x in RR`