Evaluate the following integrals:

Assume $\log x=t$

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

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

Substituting $t$ and dt in above equation we get

$\Rightarrow \int t . d t$

$\Rightarrow \frac{t^{2}}{2}+c$

But $t=\log (x)$

$\Rightarrow \frac{\log ^{2} x}{2}+C$