Question

Evaluate: ${\int }_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right|dx$

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

Answer 1

Answer: (B)

$\frac{5}{2}$

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

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

Take $z=\log x$ $⟹dz=\frac{1}{x}dx$

When $x=e^{-1},z=-1$ and when $x=1,z=0$ and when $x=e^2,z=2$

Hence, integration becomes:-

${\int }_{-1}^0-zdz+{\int }_0^{2}zdz$

$=\frac{-1}{2}{\left[z^2\right]}_{-1}^0+\frac{1}{2}{\left[z^2\right]}_0^2$

$=\frac{1}{2}+2=\frac{5}{2}$