<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] $$
\int\dfrac{1}{t^2 +a^2}\,dt=\dfrac {tan^{-1}t}{a}
$$
Therefore,
$$
\int\dfrac{12}{t^2 +3}\,dt=4\sqrt{3}{tan^{-1}\dfrac{t}{\sqrt 3}}
$$
Substitute the values.
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