1. **State the problem:** Evaluate the integral $$\int_1^2 \frac{16x^4}{(2x-1)^2} \, dx$$. 2. **Rewrite the integral:** The integral is $$\int_1^2 \frac{16x^4}{(2x-1)^2} \, dx$$. 3. **Substitution:** Let $$u = 2x - 1$$, then $$du = 2 dx$$ or $$dx = \frac{du}{2}$$. 4. **Change limits:** When $$x=1$$, $$u=2(1)-1=1$$; when $$x=2$$, $$u=2(2)-1=3$$. 5. **Express $$x$$ in terms of $$u$$:** $$x = \frac{u+1}{2}$$. 6. **Rewrite the integral in terms of $$u$$:** $$\int_1^3 \frac{16 \left(\frac{u+1}{2}\right)^4}{u^2} \cdot \frac{du}{2} = \int_1^3 \frac{16 \cdot \frac{(u+1)^4}{2^4}}{u^2} \cdot \frac{du}{2} = \int_1^3 \frac{16 (u+1)^4}{16 u^2} \cdot \frac{du}{2} = \int_1^3 \frac{(u+1)^4}{u^2} \cdot \frac{du}{2} = \frac{1}{2} \int_1^3 \frac{(u+1)^4}{u^2} du$$ 7. **Expand numerator:** $$(u+1)^4 = u^4 + 4u^3 + 6u^2 + 4u + 1$$ 8. **Rewrite integral:** $$\frac{1}{2} \int_1^3 \frac{u^4 + 4u^3 + 6u^2 + 4u + 1}{u^2} du = \frac{1}{2} \int_1^3 \left(u^2 + 4u + 6 + \frac{4}{u} + \frac{1}{u^2}\right) du$$ 9. **Integrate term-by-term:** $$\int u^2 du = \frac{u^3}{3}$$ $$\int 4u du = 2u^2$$ $$\int 6 du = 6u$$ $$\int \frac{4}{u} du = 4 \ln|u|$$ $$\int \frac{1}{u^2} du = \int u^{-2} du = -u^{-1} = -\frac{1}{u}$$ 10. **Combine integrals:** $$\frac{1}{2} \left[ \frac{u^3}{3} + 2u^2 + 6u + 4 \ln|u| - \frac{1}{u} \right]_1^3$$ 11. **Evaluate at limits:** At $$u=3$$: $$\frac{3^3}{3} + 2(3^2) + 6(3) + 4 \ln 3 - \frac{1}{3} = \frac{27}{3} + 18 + 18 + 4 \ln 3 - \frac{1}{3} = 9 + 18 + 18 + 4 \ln 3 - \frac{1}{3} = 45 + 4 \ln 3 - \frac{1}{3}$$ At $$u=1$$: $$\frac{1^3}{3} + 2(1^2) + 6(1) + 4 \ln 1 - \frac{1}{1} = \frac{1}{3} + 2 + 6 + 0 - 1 = \frac{1}{3} + 7 = \frac{22}{3}$$ 12. **Subtract:** $$\left(45 + 4 \ln 3 - \frac{1}{3}\right) - \frac{22}{3} = 45 + 4 \ln 3 - \frac{1}{3} - \frac{22}{3} = 45 + 4 \ln 3 - 8 = 37 + 4 \ln 3$$ 13. **Multiply by $$\frac{1}{2}$$:** $$\frac{1}{2} (37 + 4 \ln 3) = \frac{37}{2} + 2 \ln 3$$ **Final answer:** $$\boxed{\frac{37}{2} + 2 \ln 3}$$