<s>[INST] 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?
 [/INST] You are doing just fine. You completed the square, which is the key step, and ended up with
$$\int \frac{dx}{\sqrt{4-(x-3)^2}}.$$
Now you have a formula that you want to use. It may be better not to try to remember too many formulas. We will instead find a substitution that gives us a very familiar integral.
The above integral looks kind of like
$$\int \frac{dw}{\sqrt{1-w^2}}.$$
We look for a substitution that brings out the resemblance. 
Suppose that we let $x-3=2u$. Then the bottom will look like $\sqrt{4-4u^2}$, which is good, because we can then "take the $4$'s out."
So let's do it. Let $x-3=2u$, or if you prefer, let $(x-3)/2=u$.  Then $dx=2\,du$ and our integral becomes 
$$\int\frac{2u}{\sqrt{4-4u^2}}.$$
But $\sqrt{4-4u^2}=2\sqrt{1-u^2}$. So our integral simplifies to
$$\int \frac{du}{\sqrt{1-u^2}}=\arcsin u +C.$$
Finally, replace $u$ by $(x-3)/2$.
Another way to get to the right substitution is to note that 
$$\sqrt{4-(x-3)^2}=2\sqrt{1-((x-3)/2)^2},$$
 which makes letting $u=(x-3)/2$ very natural.  
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