Question

Find the value of \[\int {\dfrac{1}{{x\log x\log \left( {\log x} \right)}}} dx\].
A. \[\left[ {\log \left( {\log x} \right)} \right]\]
B. \[\log \left[ {\log \left( {\log x} \right)} \right]\]
C. \[\log \left[ {\log \left( {\log {x^2}} \right)} \right]\]
D. \[ - \log \left[ {\log \left( {\log x} \right)} \right]\]

Answer 1

Hint: This problem can be solved by using a substitution method. According to the substitution method, the given integral can be transformed into another form by changing the independent variable \[x{\text{ to }}t\]. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

Let \[I = \int {\dfrac{1}{{x\log x\log \left( {\log x} \right)}}} dx\]
Put \[\log \left( {\log x} \right) = t\]
Differentiating \[\log \left( {\log x} \right) = t\] w.r.t \[x\], we have
\[
   \Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right] = \dfrac{d}{{dx}}\left( t \right) \\
   \Rightarrow \dfrac{1}{{\log x}} \times \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{{dt}}{{dx}}{\text{ }}\left[ {\because \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}} \right] \\
   \Rightarrow \dfrac{1}{{x\log x}} = \dfrac{{dt}}{{dx}} \\
  \therefore dx = x\log xdt....................................................\left( 1 \right) \\
\]
Using equation (1), we get
\[
  I = \int {\dfrac{1}{{x\log x\left( t \right)}}x\log xdt} \\
  I = \int {\dfrac{1}{t}dt} \\
  I = \log t + c{\text{ }}\left[ {\because \int {\dfrac{1}{x}dx = \log x + c} } \right] \\
\]
Substituting \[\log \left( {\log x} \right) = t\], we get
\[I = \log \left[ {\log \left( {\log x} \right)} \right] + c\]
Hence, \[I = \int {\dfrac{1}{{x\log x\log \left( {\log x} \right)}}} dx = \log \left[ {\log \left( {\log x} \right)} \right] + c\]
Thus, the correct option is B. \[\log \left[ {\log \left( {\log x} \right)} \right]\]

Note: A constant namely integrating constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant. So, it is necessary to add integrating constants after completion of the integral.