1. We are asked to evaluate the integral $$\int \frac{x \, dx}{(3 - 2x - x^2)^{3/2}}.$$\n\n2. First, rewrite the quadratic expression in the denominator to a more recognizable form by completing the square:\n$$3 - 2x - x^2 = 3 - 2x - x^2 = -(x^2 + 2x - 3) = -(x^2 + 2x - 3).$$\nComplete the square inside the parentheses:\n$$x^2 + 2x - 3 = (x^2 + 2x + 1) - 1 - 3 = (x + 1)^2 - 4.$$\nSo the denominator becomes:\n$$-(x + 1)^2 + 4 = 4 - (x + 1)^2.$$\nTherefore, the integral is\n$$\int \frac{x \, dx}{(4 - (x + 1)^2)^{3/2}}.$$\n\n3. Use the substitution $$t = x + 1 \implies x = t - 1, \quad dx = dt.$$\nThe integral becomes\n$$\int \frac{(t - 1) \, dt}{(4 - t^2)^{3/2}} = \int \frac{t \, dt}{(4 - t^2)^{3/2}} - \int \frac{dt}{(4 - t^2)^{3/2}}.$$\n\n4. Evaluate each integral separately.\n\nFor the first integral:\n$$I_1 = \int \frac{t \, dt}{(4 - t^2)^{3/2}}.$$\nUse substitution $$u = 4 - t^2 \implies du = -2t \, dt \implies t \, dt = -\frac{du}{2}.$$\nSo\n$$I_1 = \int \frac{t \, dt}{u^{3/2}} = \int \frac{-\frac{du}{2}}{u^{3/2}} = -\frac{1}{2} \int u^{-3/2} \, du.$$\nIntegrate:\n$$\int u^{-3/2} \, du = \frac{u^{-1/2}}{-1/2} = -2 u^{-1/2} + C.$$\nTherefore,\n$$I_1 = -\frac{1}{2} (-2 u^{-1/2}) + C = u^{-1/2} + C = \frac{1}{\sqrt{4 - t^2}} + C.$$\n\n5. For the second integral:\n$$I_2 = \int \frac{dt}{(4 - t^2)^{3/2}}.$$\nUse the standard integral formula:\n$$\int \frac{dx}{(a^2 - x^2)^{3/2}} = \frac{x}{a^2 \sqrt{a^2 - x^2}} + C,$$\nwhere $$a = 2.$$\nSo\n$$I_2 = \frac{t}{4 \sqrt{4 - t^2}} + C.$$\n\n6. Combine the results:\n$$\int \frac{x \, dx}{(3 - 2x - x^2)^{3/2}} = I_1 - I_2 + C = \frac{1}{\sqrt{4 - t^2}} - \frac{t}{4 \sqrt{4 - t^2}} + C = \frac{4 - t}{4 \sqrt{4 - t^2}} + C.$$\nSubstitute back $$t = x + 1$$:\n$$\boxed{\int \frac{x \, dx}{(3 - 2x - x^2)^{3/2}} = \frac{4 - (x + 1)}{4 \sqrt{4 - (x + 1)^2}} + C = \frac{3 - x}{4 \sqrt{4 - (x + 1)^2}} + C}.$$