1. **State the problem:** We want to find the integral $$\int \log(x) \, dx$$. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = \log(x) \quad \Rightarrow \quad du = \frac{1}{x} dx$$ $$dv = dx \quad \Rightarrow \quad v = x$$ 4. **Apply the formula:** $$\int \log(x) \, dx = x \log(x) - \int x \cdot \frac{1}{x} \, dx = x \log(x) - \int 1 \, dx$$ 5. **Integrate the remaining integral:** $$\int 1 \, dx = x$$ 6. **Write the final answer:** $$\int \log(x) \, dx = x \log(x) - x + C$$ This means the integral of the logarithm function is $$x \log(x) - x$$ plus a constant of integration.