1. **State the problem:** We need to evaluate the integral $$\int \frac{u}{u^2 - 2u + 1} \, du$$. 2. **Simplify the denominator:** Notice that $$u^2 - 2u + 1 = (u - 1)^2$$. 3. **Rewrite the integral:** $$\int \frac{u}{(u - 1)^2} \, du$$. 4. **Use substitution:** Let $$t = u - 1$$, so $$u = t + 1$$ and $$du = dt$$. 5. **Rewrite the integral in terms of $$t$$:** $$\int \frac{t + 1}{t^2} \, dt = \int \left( \frac{t}{t^2} + \frac{1}{t^2} \right) dt = \int \left( \frac{1}{t} + t^{-2} \right) dt$$. 6. **Integrate term-by-term:** $$\int \frac{1}{t} dt = \ln|t|$$ $$\int t^{-2} dt = \int t^{-2} dt = -t^{-1} = -\frac{1}{t}$$ 7. **Combine results:** $$\ln|t| - \frac{1}{t} + C$$. 8. **Back-substitute $$t = u - 1$$:** $$\ln|u - 1| - \frac{1}{u - 1} + C$$. **Final answer:** $$\int \frac{u}{u^2 - 2u + 1} \, du = \ln|u - 1| - \frac{1}{u - 1} + C$$.