1. **State the problem:** We need to find the integral $$\int 2x \sqrt{x-1} \, dx$$ using substitution. 2. **Choose 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:** $$2(u+1) \sqrt{u} = 2(u+1) u^{\frac{1}{2}} = 2(u^{\frac{3}{2}} + u^{\frac{1}{2}})$$ 5. **Split the integral:** $$\int 2(u^{\frac{3}{2}} + u^{\frac{1}{2}}) \, du = 2 \int u^{\frac{3}{2}} \, du + 2 \int u^{\frac{1}{2}} \, du$$ 6. **Integrate each term:** $$\int u^{\frac{3}{2}} \, du = \frac{u^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} u^{\frac{5}{2}}$$ $$\int u^{\frac{1}{2}} \, du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} u^{\frac{3}{2}}$$ 7. **Combine results:** $$2 \times \frac{2}{5} u^{\frac{5}{2}} + 2 \times \frac{2}{3} u^{\frac{3}{2}} = \frac{4}{5} u^{\frac{5}{2}} + \frac{4}{3} u^{\frac{3}{2}} + C$$ 8. **Substitute back $$u = x - 1$$:** $$\frac{4}{5} (x-1)^{\frac{5}{2}} + \frac{4}{3} (x-1)^{\frac{3}{2}} + C$$ **Final answer:** $$\int 2x \sqrt{x-1} \, dx = \frac{4}{5} (x-1)^{\frac{5}{2}} + \frac{4}{3} (x-1)^{\frac{3}{2}} + C$$