<s>[INST] Calculate $\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$ $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$$
Assume that $x^2 - 6x + 12 = (x - 3)(x - 3) + 3 = (x - 3)^2 + 3$,
then $t = x - 3 \rightarrow dt = dx$,
since $$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx} = \int_{0}^{3} \frac{12}{t^2 + 3}\,dt$$.
However, I am unsure as to how to continue.
 [/INST] Hint
Your work is correct, and now you have:
$$\int{\dfrac{12}{t^{2} + 3} \,dt} = \int{\dfrac{12}{3(\frac{t^{2}}{3} + 1)} \,dt} = 4 \int{\dfrac{1}{\frac{t^{2}}{3} + 1} \,dt} = \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\sqrt{48}$$
</s>