1. **State the problem:** Evaluate the integral $$\int \frac{-9x^2}{\sqrt{4x - x^2}} \, dx.$$\n\n2. **Rewrite the expression under the square root:** Note that $$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -(x-2)^2 + 4.$$ So, $$\sqrt{4x - x^2} = \sqrt{4 - (x-2)^2}.$$ This represents a semicircle of radius 2 centered at $x=2$.\n\n3. **Use substitution:** Let $$t = x - 2,$$ so $$x = t + 2$$ and $$dx = dt.$$ The integral becomes\n$$\int \frac{-9(t+2)^2}{\sqrt{4 - t^2}} \, dt.$$\n\n4. **Expand the numerator:**\n$$(t+2)^2 = t^2 + 4t + 4,$$ so the integral is\n$$\int \frac{-9(t^2 + 4t + 4)}{\sqrt{4 - t^2}} \, dt = -9 \int \frac{t^2 + 4t + 4}{\sqrt{4 - t^2}} \, dt.$$\n\n5. **Split the integral:**\n$$-9 \left( \int \frac{t^2}{\sqrt{4 - t^2}} \, dt + 4 \int \frac{t}{\sqrt{4 - t^2}} \, dt + 4 \int \frac{1}{\sqrt{4 - t^2}} \, dt \right).$$\n\n6. **Evaluate each integral separately:**\n\n- For $$\int \frac{t^2}{\sqrt{4 - t^2}} \, dt,$$ use substitution $$t = 2\sin \theta,$$ so $$dt = 2\cos \theta \, d\theta$$ and $$\sqrt{4 - t^2} = 2\cos \theta.$$ Then,\n$$\int \frac{t^2}{\sqrt{4 - t^2}} \, dt = \int \frac{4 \sin^2 \theta}{2 \cos \theta} \cdot 2 \cos \theta \, d\theta = \int 4 \sin^2 \theta \, d\theta.$$\nUse the identity $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$ to get\n$$\int 4 \sin^2 \theta \, d\theta = 4 \int \frac{1 - \cos 2\theta}{2} \, d\theta = 2 \int (1 - \cos 2\theta) \, d\theta = 2 \left( \theta - \frac{\sin 2\theta}{2} \right) + C = 2\theta - \sin 2\theta + C.$$\n\n- For $$\int \frac{t}{\sqrt{4 - t^2}} \, dt,$$ use substitution $$u = 4 - t^2,$$ so $$du = -2t \, dt,$$ hence $$t \, dt = -\frac{du}{2}.$$ The integral becomes\n$$\int \frac{t}{\sqrt{4 - t^2}} \, dt = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2 u^{1/2} + C = -\sqrt{4 - t^2} + C.$$\n\n- For $$\int \frac{1}{\sqrt{4 - t^2}} \, dt,$$ this is a standard integral:\n$$\int \frac{1}{\sqrt{a^2 - t^2}} \, dt = \arcsin \frac{t}{a} + C,$$ so here\n$$\int \frac{1}{\sqrt{4 - t^2}} \, dt = \arcsin \frac{t}{2} + C.$$\n\n7. **Combine all results:**\n$$-9 \left( 2\theta - \sin 2\theta - 4 \sqrt{4 - t^2} + 4 \arcsin \frac{t}{2} \right) + C.$$\n\n8. **Rewrite in terms of $t$ and then $x$:**\nRecall $$t = x - 2$$ and $$\theta = \arcsin \frac{t}{2} = \arcsin \frac{x-2}{2}.$$ Also, $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{t}{2} \cdot \frac{\sqrt{4 - t^2}}{2} = \frac{t \sqrt{4 - t^2}}{2}.$$\nSo,\n$$-9 \left( 2 \arcsin \frac{x-2}{2} - \frac{(x-2) \sqrt{4 - (x-2)^2}}{2} - 4 \sqrt{4 - (x-2)^2} + 4 \arcsin \frac{x-2}{2} \right) + C.$$\nCombine the arcsin terms:\n$$2 \arcsin \frac{x-2}{2} + 4 \arcsin \frac{x-2}{2} = 6 \arcsin \frac{x-2}{2}.$$\nCombine the square root terms:\n$$- \frac{(x-2) \sqrt{4 - (x-2)^2}}{2} - 4 \sqrt{4 - (x-2)^2} = - \sqrt{4 - (x-2)^2} \left( \frac{x-2}{2} + 4 \right).$$\n\n9. **Final answer:**\n$$\boxed{ -9 \left( 6 \arcsin \frac{x-2}{2} - \sqrt{4 - (x-2)^2} \left( \frac{x-2}{2} + 4 \right) \right) + C }.$$