Answer
$$\int\frac{\log_{10}x}{x}dx=\frac{(\ln x)^2}{2\ln10}+C$$
Work Step by Step
$$A=\int\frac{\log_{10}x}{x}dx=\int\frac{\frac{\ln x}{\ln10}}{x}dx$$ $$A=\frac{1}{\ln10}\int\frac{\ln x}{x}dx$$
We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$
Therefore, $$A=\frac{1}{\ln10}\int udu$$ $$A=\frac{1}{\ln10}\times\frac{u^2}{2}+C$$ $$A=\frac{u^2}{2\ln10}+C=\frac{(\ln x)^2}{2\ln10}+C$$
