Question

The value of the integral $I={\int }_{e^{-1}}^{e^2}|\frac{\log x}{x}|dx$ is

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

Answer 1

The correct option is A $3/2$

${\int }_{e^{-1}}^{e^2}\left|\begin{array}{c}\log x\\ x\end{array}\right|dx$

Let $\log x=t$ $t=\log e^2=2$

$\frac{1}{x}dt=dt$ $t=\log e^{-1}=1$

so

${\int }_{-1}^{2}tdt$

${\left[\frac{t^2}{2}\right]}_{-1}^2$

$=\frac{2^2}{2}-\frac{(-1{)}^2}{2}$

$=2-\frac{1}{2}$

$=\frac{4-1}{2}$

$=\frac{3}{2}$.