Question

The value of $\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx$ is
$
  \left( {\text{A}} \right)\dfrac{3}{2} \\
  \left( {\text{B}} \right)\dfrac{5}{2} \\
  \left( {\text{C}} \right)3 \\
  \left( {\text{D}} \right)5 \\
 $

Answer 1

Hint: Since we have been given the function in mod, try to find the values of integral when the mod is removed for the given integral values. Try doing this by splitting the integral value. This will help us get rid of the mod. After this try to solve both the integral by using general integration rules. In this case you can use the substitution method.

Complete step-by-step solution:
The question given to us is $\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx$
We will first try to split the integral to remove the mod
It then becomes as follows
$\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \int\limits_{{e^{ - 1}}}^1 {\dfrac{{\log x}}{x}} dx + \int\limits_1^{{e^2}} {\dfrac{{\log x}}{x}} dx - - - \left( 1 \right)$
Now try to solve the integral using basic rules of integration
We start by following
We will try to substitute
$
   \Rightarrow \log x = t \\
   \Rightarrow \dfrac{d}{{dx}}\log x = \dfrac{{dt}}{{dx}} \\
   \Rightarrow \dfrac{1}{x}dx = dt \\
 $
We will try to substitute the above value in equation (1) we get
$ \Rightarrow \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \int\limits_{{e^{ - 1}}}^1 {t.dt} + \int\limits_1^{{e^2}} {t.dt} dx$
On integrating we get
$ \Rightarrow \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \left[ {\dfrac{{{t^2}}}{2}} \right]_{{e^{ - 1}}}^1 + \left[ {\dfrac{{{t^2}}}{2}} \right]_1^{{e^2}}$
Resubstituting the value of t we get
$ \Rightarrow \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \left[ {\dfrac{{{{\left( {\log x} \right)}^2}}}{2}} \right]_{{e^{ - 1}}}^1 + \left[ {\dfrac{{{{\left( {\log x} \right)}^2}}}{2}} \right]_1^{{e^2}}$
Solving further we get
$
 \Rightarrow \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \left[ {\dfrac{{{{\left( {\log 1} \right)}^2}}}{2} - \dfrac{{{{\left( {\log {e^{ - 1}}} \right)}^2}}}{2}} \right] + \left[ {\dfrac{{{{\left( {\log {e^2}} \right)}^2}}}{2} - \dfrac{{{{\left( {\log 1} \right)}^2}}}{2}} \right] \\
  \Rightarrow \left[ {\dfrac{{{{\left( {\log {e^2}} \right)}^2}}}{2} - \dfrac{{{{\left( {\log {e^{ - 1}}} \right)}^2}}}{2}} \right] \\
  \Rightarrow \dfrac{1}{2}\left[ {{{\left( {2\log e} \right)}^2} - {{\left( { - 1\log e} \right)}^2}} \right] \\
 $
Since $\log e = 1$, substituting we get
$
 \Rightarrow \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \dfrac{1}{2}\left[ {{{\left( 2 \right)}^2} - {{\left( { - 1} \right)}^2}} \right] \\
 \Rightarrow \dfrac{1}{2}\left( {4 - 1} \right) \\
  \Rightarrow \dfrac{3}{2} \\
 $
hence the value of $\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \dfrac{3}{2}$$\int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{\log x}}{x}} \right|} dx = \dfrac{3}{2}$

Hence option (A) is the correct answer.

Note: Whenever a question is given in the modulus and we are told to integrate it, as in this question, try to first calculate the value of the functions in the modulus by substituting the integral values. If you are getting both positive and negative values substitute the integral and its value where one integral term will have positive values and the second integral will have a negative value. Further solve the equation by regular integration methods. This will help us get the required answer easily.