Question

If \[{{\mathop{\rm l}\nolimits} ^r}\left( x \right)\] means \[\log \log \log \ldots \ldots \ldots x\], the log being repeated \[r\] times, then \[\int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} \] is equal to
(a) \[{{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right) + C\]
(b) \[\dfrac{{{{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right)}}{{r + 1}} + C\]
(c) \[{{\mathop{\rm l}\nolimits} ^r}\left( x \right) + C\]
(d) None of these

Answer 1

Hint:
Here, we will use the concept of integration using a substitution method. Integration by substitution method helps in transforming a given integral into simple form by substituting an independent variable. We will introduce an independent variable and equate it with the given function. We will then differentiate it with respect to \[x\] and apply the given condition to find the answer.

Complete step by step solution:
We will use substitution to integrate the given function.
Let \[t = {{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right)\].
Thus, we get
\[ \Rightarrow t = \log \log \log \ldots \ldots \ldots x\]
Here, the log is repeated \[r + 1\] times.
Differentiating both sides of the function with respect to \[x\], we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \dfrac{{d\left( {\log \log \ldots \ldots \ldots x} \right)}}{{dx}}\]
Here, the log in the denominator is repeated \[r\] times, and the log in the numerator of the second expression is also repeated \[r\] times.
Differentiating the second expression, we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \dfrac{{d\left( {\log \log \ldots \ldots \ldots x} \right)}}{{dx}}\]
Here, the log in the denominator of first expression is repeated \[r\] times, and the log in the denominator of second expression is repeated \[r - 1\] times.
This will keep on repeating due to chain rule of differentiation, with the final derivative in the chain being \[\dfrac{{d\left( {\log \left( x \right)} \right)}}{{dx}}\].
Therefore, rewriting the equation, we get
\[\begin{array}{l} \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \ldots \ldots \ldots \times \dfrac{1}{{\log \log x}} \times \dfrac{1}{{\log x}} \times \dfrac{{d\left( {\log x} \right)}}{{dx}}\\ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \dfrac{1}{{\log \log \ldots \ldots \ldots x}} \times \ldots \ldots \ldots \times \dfrac{1}{{\log \log x}} \times \dfrac{1}{{\log x}} \times \dfrac{1}{x}\end{array}\]
We know that \[{{\mathop{\rm l}\nolimits} ^r}\left( x \right)\] means \[\log \log \log \ldots \ldots \ldots x\], the log being repeated \[r\] times.
Using this formula to rewrite the expression in the integral, we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{{{\mathop{\rm l}\nolimits} ^r}\left( x \right)}} \times \dfrac{1}{{{{\mathop{\rm l}\nolimits} ^{r - 1}}\left( x \right)}} \times \ldots \ldots \ldots \times \dfrac{1}{{{{\mathop{\rm l}\nolimits} ^2}\left( x \right)}} \times \dfrac{1}{{{{\mathop{\rm l}\nolimits} ^1}\left( x \right)}} \times \dfrac{1}{x}\]
Multiplying the terms of the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{x{{\mathop{\rm l}\nolimits} ^1}\left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)}}\\ \Rightarrow \dfrac{{dt}}{{dx}} = {\left[ {x{{\mathop{\rm l}\nolimits} ^1}\left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]^{ - 1}}\\ \Rightarrow dt = {\left[ {x{{\mathop{\rm l}\nolimits} ^1}\left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]^{ - 1}}dx\end{array}\]
Now, substituting \[t = {{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right)\] and \[{\left[ {x{{\mathop{\rm l}\nolimits} ^1}\left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]^{ - 1}}dx = dt\] in the integral \[\int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} \], we get
\[\begin{array}{l}\int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} = \int {dt} \\ \Rightarrow \int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} = \int {\left( 1 \right)dt} \end{array}\]
Integrating the constant with respect to \[t\], we get
\[ \Rightarrow \int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} = t + C\]
Substituting \[t = {{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right)\] in the expression, we get the value of the integral as
\[ \Rightarrow \int {{{\left[ {x{\mathop{\rm l}\nolimits} \left( x \right){{\mathop{\rm l}\nolimits} ^2}\left( x \right){{\mathop{\rm l}\nolimits} ^3}\left( x \right) \ldots \ldots \ldots {{\mathop{\rm l}\nolimits} ^r}\left( x \right)} \right]}^{ - 1}}dx} = {{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right) + C\]
Hence, the value of the integral is \[{{\mathop{\rm l}\nolimits} ^{r + 1}}\left( x \right) + C\].

\[\therefore\] The correct option is option (a).

Note:
We should be careful while differentiating the equation \[t = \log \log \log \ldots \ldots \ldots x\] using the chain rule. We should remember the number of times log repeats in every \[\log \log \log \ldots \ldots \ldots x\]. This will help us to convert \[\log \log \log \ldots \ldots \ldots x\] to \[{{\mathop{\rm l}\nolimits} ^r}\left( x \right)\] and simplify the derivative. Here we have solved the question using integration by substitution method. This method is used to simplify a large function.