What is the value of the integral \[\int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} \]?
A. \[1 - \dfrac{1}{e}\]
B. \[2\left( {1 - \dfrac{1}{e}} \right)\]
C. \[{e^{ - 1}} - 1\]
D. None of these

Hint: Here, a definite integral is given. First, check the behaviour of the function present in the given integral \[\left| {\log x} \right|\] in the interval and find the point where the function changes it’s direction. Then, apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\] and solve the integrals to get the required answer.

Formula Used: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\], where \[c \in \left( {a,b} \right)\]
\[\int {\log xdx = x\log x - x} \]

Complete step by step solution: The given definite integral is \[\int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} \].
We know that, the log function is undefined for the negative number and it is a decreasing function in the interval \[\left[ {0,1} \right]\]. Also, \[\log \left( 1 \right) = 0\].
Means, the given function \[\log x\] changes the direction at 1.
So, apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\] and split the interval into two parts.
We get,
\[\int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = \int\limits_{\dfrac{1}{e}}^1 { - \log xdx} + \int\limits_1^e {\log xdx} \]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = - \int\limits_{\dfrac{1}{e}}^1 {\log xdx} + \int\limits_1^e {\log xdx} \]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = - \left[ {x\log x - x} \right]_{\dfrac{1}{e}}^1 + \left[ {x\log x - x} \right]_1^e\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = - \left[ {\log \left( 1 \right) - 1 - \left( {\dfrac{1}{e}\log \left( {\dfrac{1}{e}} \right) - \dfrac{1}{e}} \right)} \right] + \left[ {e\log e - e - \left( {\log \left( 1 \right) - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = - \left[ {0 - 1 - \left( {\dfrac{1}{e}\left( { - 1} \right) - \dfrac{1}{e}} \right)} \right] + \left[ {e - e - \left( {0 - 1} \right)} \right]\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = - \left[ { - 1 - \left( { - \dfrac{1}{e} - \dfrac{1}{e}} \right)} \right] + 1\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = 1 - \dfrac{2}{e} + 1\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = 2 - \dfrac{2}{e}\]
\[ \Rightarrow \int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} = 2\left( {1 - \dfrac{1}{e}} \right)\]

Option ‘B’ is correct

Note: Students often get confused about the behaviour of log function and consider it as an increasing function from the point 0. Because of that, they directly calculate \[\int\limits_{\dfrac{1}{e}}^e {\left| {\log x} \right|dx} \] as \[\int\limits_{\dfrac{1}{e}}^e {\log xdx} \] without changing the limits.