1. **Problem statement:** Evaluate the integral $$\int \frac{x \, dx}{(3 - 2x - x^2)^{3/2}}.$$\n\n2. **Rewrite the denominator:** The expression inside the power is $$3 - 2x - x^2.$$ Rewrite it in a standard quadratic form: $$-(x^2 + 2x - 3) = -(x^2 + 2x - 3).$$\n\n3. **Complete the square:** For the quadratic $$x^2 + 2x - 3,$$ complete the square:\n$$x^2 + 2x - 3 = (x^2 + 2x + 1) - 1 - 3 = (x + 1)^2 - 4.$$\nSo the denominator becomes $$-( (x + 1)^2 - 4 ) = 4 - (x + 1)^2.$$\n\n4. **Substitute:** Let $$t = x + 1,$$ so $$x = t - 1$$ and $$dx = dt.$$ The integral becomes\n$$\int \frac{(t - 1) dt}{(4 - t^2)^{3/2}}.$$\n\n5. **Split the integral:**\n$$\int \frac{t dt}{(4 - t^2)^{3/2}} - \int \frac{dt}{(4 - t^2)^{3/2}}.$$\n\n6. **Evaluate the first integral:** Use substitution $$u = 4 - t^2,$$ so $$du = -2t dt,$$ hence $$t dt = -\frac{du}{2}.$$\nThe first integral becomes\n$$\int \frac{t dt}{(4 - t^2)^{3/2}} = \int \frac{-\frac{du}{2}}{u^{3/2}} = -\frac{1}{2} \int u^{-3/2} du.$$\n\n7. **Integrate:**\n$$\int u^{-3/2} du = \int u^{-\frac{3}{2}} du = \frac{u^{-1/2}}{-1/2} = -2 u^{-1/2} + C.$$\nSo the first integral is\n$$-\frac{1}{2} \times (-2 u^{-1/2}) = u^{-1/2} = \frac{1}{\sqrt{u}} = \frac{1}{\sqrt{4 - t^2}}.$$\n\n8. **Evaluate the second integral:**\n$$\int \frac{dt}{(4 - t^2)^{3/2}}.$$\nUse the standard integral formula:\n$$\int \frac{dx}{(a^2 - x^2)^{3/2}} = \frac{x}{a^2 \sqrt{a^2 - x^2}} + C,$$ where $$a=2.$$\nSo\n$$\int \frac{dt}{(4 - t^2)^{3/2}} = \frac{t}{4 \sqrt{4 - t^2}} + C.$$\n\n9. **Combine results:**\n$$\int \frac{(t - 1) dt}{(4 - t^2)^{3/2}} = \frac{1}{\sqrt{4 - t^2}} - \frac{t}{4 \sqrt{4 - t^2}} + C = \frac{4 - t}{4 \sqrt{4 - t^2}} + C.$$\n\n10. **Back-substitute:** Recall $$t = x + 1,$$ so\n$$\int \frac{x \, dx}{(3 - 2x - x^2)^{3/2}} = \frac{4 - (x + 1)}{4 \sqrt{4 - (x + 1)^2}} + C = \frac{3 - x}{4 \sqrt{4 - (x + 1)^2}} + C.$$