1. **State the problem:** We want to find the integral $$\int \log(x) \, dx$$. 2. **Use integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \log(x)$$ and $$dv = dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{1}{x} dx$$ - $$v = \int dx = x$$ 5. **Apply the formula:** $$\int \log(x) \, dx = x \log(x) - \int x \cdot \frac{1}{x} \, dx = x \log(x) - \int 1 \, dx$$ 6. **Integrate the remaining integral:** $$\int 1 \, dx = x$$ 7. **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.