1. **State the problem:** We need to find the integral of the function $$\frac{1}{x} + \frac{5}{(x-2)^2}$$ with respect to $x$. 2. **Rewrite the integral:** $$\int \left( \frac{1}{x} + \frac{5}{(x-2)^2} \right) dx = \int \frac{1}{x} dx + \int \frac{5}{(x-2)^2} dx$$ 3. **Integrate the first term:** The integral of $\frac{1}{x}$ is a standard form: $$\int \frac{1}{x} dx = \ln|x| + C_1$$ 4. **Integrate the second term:** Recall that: $$\int (x - a)^{-n} dx = \frac{(x - a)^{-n+1}}{-n+1} + C, \quad n \neq 1$$ Here, $n=2$ and $a=2$, so: $$\int \frac{5}{(x-2)^2} dx = 5 \int (x-2)^{-2} dx = 5 \cdot \frac{(x-2)^{-1}}{-1} + C_2 = -\frac{5}{x-2} + C_2$$ 5. **Combine the results:** $$\int \left( \frac{1}{x} + \frac{5}{(x-2)^2} \right) dx = \ln|x| - \frac{5}{x-2} + C$$ 6. **Final answer:** $$\boxed{\ln|x| - \frac{5}{x-2} + C}$$ This is the antiderivative of the given function.