what is ${\int }_a^b \frac{\log x}{x}dx$ equal to ?

Let the given integral be

$I={\int }_a^b \frac{\log x}{x}dx$

Put log x = t and $\frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}$ when x = a ,  t = log a and if x = b, t = log b.

$\begin{array}{l}\therefore \mathrm{I}={\int }_{\log \mathrm{a}}^{\log \mathrm{b}}t \mathrm{dt}=\left[{\left.\frac{{\mathrm{t}}^2}{2}\right|}_{\mathrm{loga}}^{\mathrm{logb}}\right.\\ =\frac{1}{2}\left[(\log \mathrm{b}{)}^2-(\log \mathrm{a}{)}^2\right]\\ =\frac{1}{2}\left[\right(\log \mathrm{b}+\log \mathrm{a}\left)\right(\log \mathrm{b}-\log \mathrm{a}\left)\right]\\ =\frac{1}{2}\log \left(\mathrm{ab}\right)\log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\end{array}$