$\int log \, x \, dx=x \,log x-x$
[By Using integration by parts]
$log x is -ve. for x, < 1$ and $+ve$ for $x > 1 $.
Also $log_{10} \,x= \frac{log\,x}{log\,10}$
$\therefore $ given integral
$=\frac{1}{log\,10}\left[\int\limits_{1 /2}^{1}-log\,x\,dx+\int\limits_{1}^{2} log\,x\,dx\right]$
$=-\frac{1}{log\,10} \left[x\,log\,x-x\right]_{1 /2}^{1}+\frac{1}{log\,10} \left[x\,log\,x-x\right]_{1}^{2}$
$=-\frac{1}{log\,10}\left[-1-\frac{1}{2}log\frac{1}{2}+\frac{1}{2}\right]+\frac{1}{log\,10} \left[2\,log\,2-2+1\right]$
$=\frac{1}{log\,10}\left[\frac{1}{2}-\frac{1}{2}log\,2\right]+\frac{2\,log\,2-1}{log\,10}$
$=\frac{4\,log\,2-2+1-log\,2}{2\,log\,10}=\frac{3\,log\,2-1}{2\,log\,10}$
$=\frac{1}{2} \frac{log\,8-log\,e}{log\,10}=\frac{1}{2} \frac{log\left(\frac{8}{e}\right)}{log\,10}$
$=\frac{1}{2}log_{10} \left(\frac{8}{e}\right)$