wat

If you didn't understand that last part you probably haven't learned about trig substitutions yet. Here is the problem worked without using a trigonometric substitution. Before the substitution we had,

∫1/sqrt(9-u^2)du, where u=x-3.

Factor 9 out of the square root.

∫1/sqrt(9(1-(1/9)u^2)du=∫1/3sqrt((1-(1/9)u^2)du=∫1/3sqrt((1-((1/3)u)^2)du

Let v=(1/3)u, then dv=(1/3)du

∫1/sqrt((1-v^2)du

Now integrate and back substitute.

sin^-1(v)+C=sin^-1((1/3)u)+C=sin^-1((1/3)(x-3))+C

I probably should have done it this way when I first saw the answer and realized it could be solved this way, but I found the solution using the trigonometric substitution more interesting.