Solution:

Let $\log x=t$
$\begin{array}{l}
\Rightarrow d(\log x)=d t \\
\Rightarrow \frac{1}{x} d x=d t
\end{array}$
By substituting $\mathrm{t}$ and $dt$ in above equation we obtain
$\begin{array}{l}
\Rightarrow \int \mathrm{t} . \mathrm{dt} \\
\Rightarrow \frac{\mathrm{t}^{2}}{2}+\mathrm{c}
\end{array}$
But $t=\log (x)$
$\Rightarrow \frac{\log ^{2} x}{2}+c$