Evaluate: ∫ log₁₀ x dx

$\log_{10} x = \frac{\ln x}{\ln 10}$, so the integral is $\frac{1}{\ln 10}\int \ln x \cdot 1\,dx$. By parts: $\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x$. Hence the integral $= \frac{x(\ln x - 1)}{\ln 10} + C = x(\log_{10} x - \log_{10} e) + C$.