1. **Problem statement:** Calculate the integral $$\int \frac{\ln x}{x} \, dx$$. 2. **Recall integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Assign parts:** Let $$u = \ln x$$ so $$du = \frac{1}{x} dx$$. Let $$dv = \frac{1}{x} dx$$ so $$v = \ln x$$. 4. **Apply integration by parts:** $$\int \frac{\ln x}{x} dx = u v - \int v \, du = (\ln x)(\ln x) - \int (\ln x) \frac{1}{x} dx = (\ln x)^2 - \int \frac{\ln x}{x} dx$$ 5. **Notice the integral on the right is the same as the original integral.** Move it to the left side: $$\int \frac{\ln x}{x} dx + \int \frac{\ln x}{x} dx = (\ln x)^2$$ 6. **Combine like terms:** $$2 \int \frac{\ln x}{x} dx = (\ln x)^2$$ 7. **Solve for the integral:** $$\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2} + C$$ **Final answer:** $$\boxed{\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2} + C}$$