1. **State the problem:** We need to evaluate the integral $$\int \sqrt{x^3 - 3} (x^2 - 1) \, dx.$$\n\n2. **Rewrite the integral:** Let $$I = \int \sqrt{x^3 - 3} (x^2 - 1) \, dx.$$\n\n3. **Substitution:** Notice the expression inside the square root is $$x^3 - 3$$ and the other factor is $$x^2 - 1$$. We try substitution $$u = x^3 - 3$$. Then, $$\frac{du}{dx} = 3x^2 \Rightarrow du = 3x^2 \, dx.$$\n\n4. **Expressing the integral in terms of $$u$$:** We have $$x^2 \, dx = \frac{du}{3}$$. The integral becomes\n$$I = \int \sqrt{u} (x^2 - 1) \, dx = \int \sqrt{u} \left(x^2 \, dx - dx\right).$$\n\n5. **Rewrite $$dx$$ in terms of $$du$$:** From $$du = 3x^2 \, dx$$, we get $$dx = \frac{du}{3x^2}$$. Substitute back:\n$$I = \int \sqrt{u} \left(x^2 \cdot dx - dx\right) = \int \sqrt{u} \left(x^2 \cdot dx\right) - \int \sqrt{u} \, dx.$$\n\n6. **Substitute for $$x^2 dx$$ and $$dx$$:**\n- $$x^2 dx = \frac{du}{3}$$\n- $$dx = \frac{du}{3x^2}$$ (but this is complicated, so instead express $$x^2$$ in terms of $$u$$).\n\n7. **Express $$x^2$$ in terms of $$u$$:** From $$u = x^3 - 3$$, we get $$x^3 = u + 3$$. Then, $$x = (u + 3)^{1/3}$$ and $$x^2 = \left((u + 3)^{1/3}\right)^2 = (u + 3)^{2/3}.$$\n\n8. **Rewrite $$dx$$:** Since $$du = 3x^2 dx$$, then $$dx = \frac{du}{3x^2} = \frac{du}{3 (u + 3)^{2/3}}.$$\n\n9. **Rewrite the integral:**\n$$I = \int \sqrt{u} \left(x^2 dx - dx\right) = \int \sqrt{u} \left(\frac{du}{3} - \frac{du}{3 (u + 3)^{2/3}}\right) = \frac{1}{3} \int u^{1/2} du - \frac{1}{3} \int \frac{u^{1/2}}{(u + 3)^{2/3}} du.$$\n\n10. **Simplify the first integral:**\n$$\frac{1}{3} \int u^{1/2} du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2}.$$\n\n11. **The second integral is complicated:**\n$$\int \frac{u^{1/2}}{(u + 3)^{2/3}} du,$$ which does not simplify nicely with elementary functions.\n\n12. **Conclusion:** The integral $$\int \sqrt{x^3 - 3} (x^2 - 1) \, dx$$ can be partially expressed as\n$$I = \frac{2}{9} (x^3 - 3)^{3/2} - \frac{1}{3} \int \frac{(x^3 - 3)^{1/2}}{(x^3)^{2/3}} du + C,$$ but the remaining integral is non-elementary and may require special functions or numerical methods.\n\n**Final answer:**\n$$\boxed{\int \sqrt{x^3 - 3} (x^2 - 1) \, dx = \frac{2}{9} (x^3 - 3)^{3/2} - \frac{1}{3} \int \frac{(x^3 - 3)^{1/2}}{(x^3)^{2/3}} du + C}.$$