Evaluate the following integrals:

Let $I=\int \log _{10} x d x$

$=\int \frac{\log x}{\log 10} d x$

$=\frac{1}{\log 10} \int 1 \times \log x d x$

Using integration by parts,

$=\frac{1}{\log 10}\left(\log x \int d x-\int \frac{d}{d x} \log x \int 1 d x\right)$

We know that $\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}=\frac{1}{\mathrm{x}}$

$=\frac{1}{\log 10}\left(x \log x-\int \frac{1}{x} x x d x\right)$

$=\frac{1}{\log 10}\left(x \log x-\int d x\right)$

$=\frac{1}{\log 10}(x \log x-x)+c$

$=\frac{x}{\log 10}(1-\log x)+c$