1. **State the problem:** Find the integral of $$\left(\frac{x}{x-2}\right)^2$$ with respect to $x$. 2. **Rewrite the integrand:** Simplify the expression inside the integral. $$\left(\frac{x}{x-2}\right)^2 = \frac{x^2}{(x-2)^2}$$ 3. **Use substitution:** Let $$u = x - 2$$, then $$x = u + 2$$ and $$dx = du$$. Rewrite the integral in terms of $u$: $$\int \frac{(u+2)^2}{u^2} du$$ 4. **Expand the numerator:** $$(u+2)^2 = u^2 + 4u + 4$$ So the integral becomes: $$\int \frac{u^2 + 4u + 4}{u^2} du = \int \left(\frac{u^2}{u^2} + \frac{4u}{u^2} + \frac{4}{u^2}\right) du = \int \left(1 + \frac{4}{u} + 4u^{-2}\right) du$$ 5. **Integrate term-by-term:** $$\int 1 \, du = u$$ $$\int \frac{4}{u} \, du = 4 \ln|u|$$ $$\int 4u^{-2} \, du = 4 \int u^{-2} \, du = 4 \left(-u^{-1}\right) = -\frac{4}{u}$$ 6. **Combine the results:** $$u + 4 \ln|u| - \frac{4}{u} + C$$ 7. **Substitute back to $x$:** $$u = x - 2$$ Final answer: $$\boxed{(x - 2) + 4 \ln|x - 2| - \frac{4}{x - 2} + C}$$