1. **Problem Statement:** Find the integral $$\int \frac{x+3}{(x-1)^2} \, dx$$ given that $$z = x - 1$$. 2. **Substitution:** Since $$z = x - 1$$, then $$x = z + 1$$. Rewrite the integral in terms of $$z$$: $$\int \frac{(z+1)+3}{z^2} \, dx = \int \frac{z + 4}{z^2} \, dx$$. 3. **Rewrite the integral:** $$\int \frac{z + 4}{z^2} \, dx = \int \left( \frac{z}{z^2} + \frac{4}{z^2} \right) \, dx = \int \left( \frac{1}{z} + \frac{4}{z^2} \right) \, dx$$. 4. **Change variable of integration:** Since $$z = x - 1$$, then $$dz = dx$$. So the integral becomes: $$\int \left( \frac{1}{z} + \frac{4}{z^2} \right) \, dz = \int \frac{1}{z} \, dz + 4 \int z^{-2} \, dz$$. 5. **Integrate each term:** - $$\int \frac{1}{z} \, dz = \ln|z| + C_1$$ - $$4 \int z^{-2} \, dz = 4 \left( -z^{-1} \right) + C_2 = -\frac{4}{z} + C_2$$ 6. **Combine results:** $$\int \frac{x+3}{(x-1)^2} \, dx = \ln|z| - \frac{4}{z} + C$$ 7. **Substitute back $$z = x - 1$$:** $$\boxed{\ln|x - 1| - \frac{4}{x - 1} + C}$$ **Answer corresponds to option (c).**