Question:

Step 1: Simplify the expression in the denominator by completing the square. The given integral is $I = \int \frac{dx}{\sqrt{12x - 9x^2}}$. We begin by rewriting the quadratic expression under the square root using the method of completing the square. $$12x - 9x^2 = -9x^2 + 12x$$ Factor out $-9$ from the terms involving $x$: $$= -9\left(x^2 - \frac{12}{9}x\right)$$ $$= -9\left(x^2 - \frac{4}{3}x\right)$$ To complete the square for $x^2 - \frac{4}{3}x$, we add and subtract $\left(\frac{-4/3}{2}\right)^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}$: $$= -9\left(x^2 - \frac{4}{3}x + \frac{4}{9} - \frac{4}{9}\right)$$ $$= -9\left(\left(x - \frac{2}{3}\right)^2 - \frac{4}{9}\right)$$ Distribute the $-9$ back into the expression: $$= -9\left(x - \frac{2}{3}\right)^2 + 9\left(\frac{4}{9}\right)$$ $$= -(3x - 2)^2 + 4$$ $$= 4 - (3x - 2)^2$$ Substitute this back into the integral: $$I = \int \frac{dx}{\sqrt{4 - (3x - 2)^2}}$$ Step 2: Apply a suitable substitution to transform the integral. To simplify the integral into a standard form, we make the substitution $t = 3x - 2$. Differentiating both sides with respect to $x$: $$\frac{dt}{dx} = 3$$ This implies $dx = \frac{dt}{3}$. Substitute $t$ and $dx$ into the integral: $$I = \int \frac{\frac{dt}{3}}{\sqrt{4 - t^2}}$$ $$I = \frac{1}{3} \int \frac{dt}{\sqrt{2^2 - t^2}}$$ Step 3: Evaluate the integral and substitute back to express the result in terms of $x$. The integral is now in the standard form $\int \frac{dy}{\sqrt{a^2 - y^2}} = \sin^{-1}\left(\frac{y}{a}\right) + C$. In our case, $a=2$ and the variable is $t$. Applying the formula: $$I = \frac{1}{3} \sin^{-1}\left(\frac{t}{2}\right) + C$$ Finally, substitute back $t = 3x - 2$ to express the result in terms of $x$: $$I = \frac{1}{3} \sin^{-1}\left(\frac{3x - 2}{2}\right) + C$$ The final answer is $\boxed{\frac{1}{3} \sin^{-1} \left( \frac{3x - 2}{2} \right) + C}$.