Integrate $\frac{1}{\sqrt{6x-x^2-5}}$ Question: Integrate $\frac{1}{\sqrt{6x-x^2-5}}$
My Working: 
$$\int{\frac{1}{\sqrt{6x-x^2-5}}} = \int{\frac{1}{\sqrt{-(x^2-6x+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-3^2+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-4)}}}$$
Is it right so far? Because I think I should be using the formula
$$\int{\frac{1}{\sqrt{a^2-(x+b)^2}}} = \sin^{-1} \left( \frac{x+b}{a} \right)+c$$
$-4$ can't fit into $a^2$ can it? I haven't covered complex numbers yet, so I must have made a mistake?

Your derivation is correct, but you aren't looking at the last line correctly. Notice that
$$-((x-3)^2-4)=4-(x-3)^2,$$
so your parameters are $a=2$ and $b=-3$ to plug into the formula. We rule out $a=-2$ because the square root function is always nonnegative, and so following holds only for $a$ positive:
$$\frac{d}{dx}\sin^{-1}\left(\frac{x+b}{a}\right)=\frac{1}{a\sqrt{1-(x+b)^2/a^2}}=\frac{1}{\sqrt{a^2-(x+b)^2}}.$$
