1. **State the problem:** We want to evaluate the integral $$\int 2x \sqrt{x-1} \, dx$$ using substitution. 2. **Choose a substitution:** Let $$u = x - 1$$. Then, $$du = dx$$ and $$x = u + 1$$. 3. **Rewrite the integral in terms of $$u$$:** $$\int 2x \sqrt{x-1} \, dx = \int 2(u+1) \sqrt{u} \, du$$ 4. **Simplify the integrand:** $$2(u+1) \sqrt{u} = 2(u+1) u^{1/2} = 2(u^{3/2} + u^{1/2})$$ 5. **Split the integral:** $$\int 2(u^{3/2} + u^{1/2}) \, du = 2 \int u^{3/2} \, du + 2 \int u^{1/2} \, du$$ 6. **Integrate each term:** $$\int u^{3/2} \, du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}$$ $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}$$ 7. **Combine results:** $$2 \times \frac{2}{5} u^{5/2} + 2 \times \frac{2}{3} u^{3/2} = \frac{4}{5} u^{5/2} + \frac{4}{3} u^{3/2} + C$$ 8. **Substitute back to $$x$$:** $$\frac{4}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + C$$ **Final answer:** $$\int 2x \sqrt{x-1} \, dx = \frac{4}{5} (x-1)^{5/2} + \frac{4}{3} (x-1)^{3/2} + C$$