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.
 A: It is $\frac{1}{3} \int{\dfrac{12}{\left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1} \,dt}$, since $\dfrac{1}{t^{2} + 3} = \dfrac{1}{3 \left(\dfrac{t^{2}}{3} + 1 \right)} = \dfrac{1}{3 \left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1}$.
A: Hint
Your work is correct, and now you have:
$$\int{\dfrac{12}{t^{2} + 3} \,dt} = \int{\dfrac{12}{3(\frac{t^{2}}{3} + 1)} \,dt} = 4 \int{\dfrac{1}{\frac{t^{2}}{3} + 1} \,dt} = \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\sqrt{48}$$
A: $$
\int\dfrac{1}{t^2 +a^2}\,dt=\dfrac {tan^{-1}t}{a}
$$
Therefore,
$$
\int\dfrac{12}{t^2 +3}\,dt=4\sqrt{3}{tan^{-1}\dfrac{t}{\sqrt 3}}
$$
Substitute the values.
A: 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
$\int\frac{1}{1+x^2}dx=\arctan x+C$
