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$$\int_{0}^{3}{\left(\frac{12}{x^2 - 6x + 12}\right) \,dx}$$

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.

Taylor
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user155971
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4 Answers4

1

It is $\frac{1}{3} \int{\dfrac{12}{\left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1} \,dt}$, since $\dfrac{1}{t^{2} + 3} = \dfrac{1}{3 \left(\dfrac{t^{2}}{3} + 1 \right)} = \dfrac{1}{3 \left(\dfrac{t}{\sqrt{3}}\right)^{2} + 1}$.

Taylor
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Dr. Sonnhard Graubner
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Hint

Your work is correct, and now you have:

$$\int{\dfrac{12}{t^{2} + 3} \,dt} = \int{\dfrac{12}{3(\frac{t^{2}}{3} + 1)} \,dt} = 4 \int{\dfrac{1}{\frac{t^{2}}{3} + 1} \,dt} = \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\sqrt{48}$$

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

vidhan
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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 $\int\frac{1}{1+x^2}dx=\arctan x+C$

mi986
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