Question

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

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $3$
• D. $5$

Answer 1

Answer: (B)

$\frac{5}{2}$

${\int }_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right|dx={\int }_{e^{-1}}^1\left|\frac{\log x}{x}\right|dx+{\int }_1^{e^2}\left|\frac{\log x}{x}\right|dx$

$=-{\int }_{e^{-1}}^1\frac{\log x}{x}dx+{\int }_1^{e^2}\frac{\log x}{x}dx$

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