My problem is: $$\int \frac{1}{(x-2)\sqrt{x^{2}-4x+3}}\, dx$$
Complete the square, $$\int \frac{1}{(x-2)\sqrt{(x-2)^{2}-1}}\, dx$$
I know I'm probably supposed to use $ \frac{d}{dx}\operatorname{arcsec}(u) = \frac{1}{|u|\sqrt{(u^2 - 1)}} * \frac{du}{dx}$ for the trig substitution
but what would that give me? Any help would be useful. I don't remember trig super well.
Use the trigonometric substitution $x=\sec{u}+2$. Then $dx=\sec{u}\tan{u}du$. Hence the integral becomes $$\int\frac{\sec{u}\tan{u}}{\sec{u}\tan{u}}du=u+c=\arccos{\left(\frac{1}{x-2}\right)}+c$$
HINT:
$$x-2=\cosh t$$
Then:
$$\sqrt{(x-2)^2-1}=\sinh t, \ \ \ dx=\sinh t dt$$
From which
$$\int \frac{1}{(x-2)\sqrt{(x-2)^{2}-1}} dx=\int \frac{1}{\cosh t \sinh t} \sinh t dt=\int \frac{1}{\cosh t} dt$$
