Question

The value of ${\int }_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right|dx$ is

• A. $3/2$
• B. $5/2$
• C. $3$
• D. $5$

Answer 1

The correct option is B $5/2$

${\int }_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right|dx={\int }_1^{e^2}\left|\frac{\log x}{x}\right|dx+{\int }_{e^{-1}}^1\left|\frac{\log x}{x}\right|dx$
$=-{\int }_{e^{-1}}^1\frac{\log x}{x}dx+{\int }_1^{e^2}\frac{\log x}{x}dx$
$=\frac{1}{2}{\left[{\left(\log x\right)}^2\right]}_1^{e^2}-\frac{1}{2}{\left[{\left(\log x\right)}^2\right]}_{e^{-1}}$
$=\frac{1}{2}(2^2-0)-\frac{1}{2}(0-1^2)=\frac{5}{2}$