$\int_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right| d x=$
$\int_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right| d x=$
$\frac{2}{5}$
2
5
$\frac{5}{2}$
Solution
$I=\int_{e^{-1}}^{e^2}\left|\frac{\log x}{x}\right| d x=\int_{e^{-1}}^1\left(-\frac{\log x}{x}\right) d x+\int_1^{e^2}\left(\frac{\log x}{x}\right) d x$
On putting $\log x=t \Rightarrow \frac{d x}{x}=d t$
$
\begin{aligned}
& =-\int_{-1}^0 t d t+\int_0^2 t d t=\left[\frac{-t^2}{2}\right]_{-1}^0+\left[\frac{t^2}{2}\right]_0^2 \\
& =-\frac{1}{2}(0-1)+\frac{1}{2}(4-0)=\frac{1}{2}+2=\frac{5}{2}
\end{aligned}
$
Hence, option (d) is correct