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. $- \log \left[ {\log \left( {\log x} \right)} \right]$

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.

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
$\Rightarrow \dfrac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right] = \dfrac{d}{{dx}}\left( t \right) \\ \Rightarrow \dfrac{1}{{\log x}} \times \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{{dt}}{{dx}}{\text{ }}\left[ {\because \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}} \right] \\ \Rightarrow \dfrac{1}{{x\log x}} = \dfrac{{dt}}{{dx}} \\ \therefore dx = x\log xdt....................................................\left( 1 \right) \\$
$I = \int {\dfrac{1}{{x\log x\left( t \right)}}x\log xdt} \\ I = \int {\dfrac{1}{t}dt} \\ I = \log t + c{\text{ }}\left[ {\because \int {\dfrac{1}{x}dx = \log x + c} } \right] \\$
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. $\log \left[ {\log \left( {\log x} \right)} \right]$