Evaluate the following integrals:

Assume $x+\log x=t$

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

$\Rightarrow\left(1+\frac{1}{x}\right) d x=d t$

$\Rightarrow\left(\frac{x+1}{x}\right) d x=d t$

Put $t$ and $d t$ in the given equation we get

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

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

But $t=x+\log x$

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