1. **State the problem:** Evaluate the definite integral $$\int_2^5 \frac{x}{\sqrt{x-1}} \, dx$$. 2. **Rewrite the integral:** Let us express the integral in a simpler form: $$\int_2^5 \frac{x}{\sqrt{x-1}} \, dx = \int_2^5 x (x-1)^{-\frac{1}{2}} \, dx$$. 3. **Use substitution:** Let $$u = x - 1$$, then $$du = dx$$ and when $$x=2$$, $$u=1$$; when $$x=5$$, $$u=4$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int_1^4 (u+1) u^{-\frac{1}{2}} \, du = \int_1^4 (u^{\frac{1}{2}} + u^{-\frac{1}{2}}) \, du$$. 5. **Split the integral:** $$\int_1^4 u^{\frac{1}{2}} \, du + \int_1^4 u^{-\frac{1}{2}} \, du$$. 6. **Integrate each term:** - For $$\int u^{\frac{1}{2}} \, du$$, use the power rule: $$\int u^{\frac{1}{2}} \, du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} u^{\frac{3}{2}}$$. - For $$\int u^{-\frac{1}{2}} \, du$$: $$\int u^{-\frac{1}{2}} \, du = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2 u^{\frac{1}{2}}$$. 7. **Evaluate the definite integrals:** $$\left[ \frac{2}{3} u^{\frac{3}{2}} \right]_1^4 + \left[ 2 u^{\frac{1}{2}} \right]_1^4 = \left( \frac{2}{3} (4)^{\frac{3}{2}} - \frac{2}{3} (1)^{\frac{3}{2}} \right) + \left( 2 (4)^{\frac{1}{2}} - 2 (1)^{\frac{1}{2}} \right)$$. 8. **Calculate powers:** - $$4^{\frac{3}{2}} = (\sqrt{4})^3 = 2^3 = 8$$. - $$1^{\frac{3}{2}} = 1$$. - $$4^{\frac{1}{2}} = 2$$. - $$1^{\frac{1}{2}} = 1$$. 9. **Substitute values:** $$\left( \frac{2}{3} \times 8 - \frac{2}{3} \times 1 \right) + \left( 2 \times 2 - 2 \times 1 \right) = \left( \frac{16}{3} - \frac{2}{3} \right) + (4 - 2) = \frac{14}{3} + 2$$. 10. **Simplify:** $$\frac{14}{3} + 2 = \frac{14}{3} + \frac{6}{3} = \frac{20}{3}$$. **Final answer:** $$\boxed{\frac{20}{3}}$$