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.

Note that with your substitution the limits of integration should be changed from -3 to 0.
\begin{align*}
\int \frac{12}{t^2 +3}dt
&=\int\frac{12}{3(\frac{t^2}{3}+1)}dt\\
&=\int\frac{12}{3(\frac{t^2}{(\sqrt{3})^2}+1)}dt\\
&=\int\frac{12}{3\left(\left(\frac{t}{\sqrt{3}}\right)^2+1\right)}dt\\
&=\frac{1}{3}\int\frac{12}{\left(\frac{t}{\sqrt{3}}\right)^2+1}dt\\
\end{align*}
Then you can apply another substitution where $u=\frac{t}{\sqrt3}$ and then use the fact that
$\int\frac{1}{1+x^2}dx=\arctan x+C$
