Integrate $\sqrt{x^2 - 2x}$ $$\int{\sqrt{x^2 - 2x}}$$
I think I should be doing trig substitution, but which? I completed the square giving 
$$\int{\sqrt{(x-1)^2 -1}}$$
But the closest I found is  for
$$\frac{1}{\sqrt{a^2 - (x+b)^2}}$$ 
So I must add a $-$, but how? 

The standard way to solve these problems is indeed with trigonometric substitutions. 
But it is not the only way to solve these. 
Another way that was used more before is called Euler substitutions

if we are to integrate $\sqrt{ax^2+bx+c}$ that has real roots $\alpha$ and
  $\beta$, then we can use the substitution $\sqrt{ax^2+bx+c}=x \cdot t$

Now, in your problem we have $\alpha=0$ and $\beta=2$, so we may choose to use the substitution 
$$ \sqrt{x^2-2x} = (x-0)\cdot t \ \Rightarrow \ x = \frac{2}{1-t^2}$$
Since this is homework and I am lazy, so from here I will only outline the details
$$ \begin{align} I & = \int \sqrt{x^2-2x}\,\mathrm{d}x \\
                   & = \int \frac{8t^2}{1-3t^2+3t^4-t^6}\,\mathrm{d}t \\
                   & = \int -\frac{8t^2}{(t+1)^3(t-1)^3}\,\mathrm{d}t \\
                   & = \int - \frac{1}{2}\frac{1}{t+1} + \frac{1}{(t+1)^3} 
- \frac{1}{2}\frac{1}{(t+1)^2} + \frac{1}{2}\frac{1}{(t-1)^2} - \frac{1}{(t-1)^3} - \frac{1}{2}\frac{1}{(t-1)^2} \, \mathrm{d}t \\
\end{align} $$ 
and from here the rest is obvious =)
