Evaluate the following integrals:

Assume $\log x=t$

$\mathrm{d}(\log x)=\mathrm{dt}$

$\frac{1}{x} d x=d t$

Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get

$\Rightarrow \int \frac{d t}{t}$

$=\ln |\mathrm{t}|+\mathrm{c}$

But $t=\log x$

$=\ln |\log x|+c$