Evaluate the following integrals:

Assume $3+\log x=t$

$\mathrm{d}(3+\log \mathrm{x})=\mathrm{dt}$

$\Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$

Put $t$ and dt in given equation we get

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

$=\ln |t|+c .$

But $t=3+\log x$

$=\ln |3+\log x|+c$