Question

Evaluate the integration $\int{\left( \log \left( \log x \right)+\dfrac{1}{{{\left( \log x \right)}^{2}}} \right)dx}$?

Answer 1

Hint: Assume the given integral as I. Now, substitute $\log x=k$ and convert it into the exponential function $x={{e}^{k}}$. Differentiate both the sides with respect to x and substitute the value of $dx$ in terms of $dk$ in the integral. Now, break the integral into two parts and leave the second part as it is. For the first part use the formula use the ILATE rule and consider $\log k$ as function 1 $\left( {{f}_{1}}\left( k \right) \right)$ and ${{e}^{k}}$ as function 2 $\left( {{f}_{2}}\left( k \right) \right)$ and apply the rule of integration by parts given as \[\int{{{f}_{1}}\left( k \right)\times {{f}_{2}}\left( k \right)dk}=\left[ {{f}_{1}}\left( k \right)\int{{{f}_{2}}\left( k \right)dk} \right]-\int{\left[ {{f}_{1}}'\left( k \right)\int{{{f}_{2}}\left( k \right)dk} \right]dk}\] to simplify. Apply the same formula and rule once more by considering $\dfrac{1}{k}$ as function 1 and ${{e}^{k}}$ as function 2. Cancel the common terms and substitute back the value of k in terms of x to get the answer.

Complete step-by-step solution:
Here we are asked to find the integral of the function $\left( \log \left( \log x \right)+\dfrac{1}{{{\left( \log x \right)}^{2}}} \right)$. Let us use the substitution method to get the answer. Assuming the integral as I we have,
$\Rightarrow I=\int{\left( \log \left( \log x \right)+\dfrac{1}{{{\left( \log x \right)}^{2}}} \right)dx}$
Substituting $\log x=k$ we can write it in the exponential form as $x={{e}^{k}}$, so differentiating both the sides to find the value of $dx$ in terms of $dk$ we get,
$\begin{align}
  & \Rightarrow dx=d\left( {{e}^{k}} \right) \\
 & \Rightarrow dx={{e}^{k}}dk \\
\end{align}$
Substituting the values in the above integral we get,
$\begin{align}
  & \Rightarrow I=\int{\left( \log k+\dfrac{1}{{{\left( k \right)}^{2}}} \right){{e}^{k}}dk} \\
 & \Rightarrow I=\int{\left( {{e}^{k}}\log k+\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}} \right)dk} \\
\end{align}$
Breaking the integral into two parts we get,
$\Rightarrow I=\int{\left( {{e}^{k}}\log k \right)dk}+\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk}$
Now, let us leave the second part of the integral as it is and evaluate the first part. Clearly we can see that in the integral $\int{{{e}^{k}}\log kdk}$ we have a product of an exponential function and a logarithmic function so we need to apply integration by parts to find the integral. Now, according to the ILATE rule we have to $\log k$ as function 1 $\left( {{f}_{1}}\left( k \right) \right)$ and ${{e}^{k}}$ as function 2 $\left( {{f}_{2}}\left( k \right) \right)$. Here, ILATE stands for: -
I – Inverse trigonometric function
L – Logarithmic function
A – Algebraic function
T – Trigonometric function
E – Exponential function
The numbering of the functions is done according to the order of appearance in the above list. Therefore we have the formula \[\int{{{f}_{1}}\left( k \right)\times {{f}_{2}}\left( k \right)dk}=\left[ {{f}_{1}}\left( k \right)\int{{{f}_{2}}\left( k \right)dk} \right]-\int{\left[ {{f}_{1}}'\left( k \right)\int{{{f}_{2}}\left( k \right)dk} \right]dk}\] to calculate the integral of product of two functions. So we get,
$\begin{align}
  & \Rightarrow I=\log k\times {{e}^{k}}-\int{{{e}^{k}}\times \dfrac{1}{k}dk}+\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk} \\
 & \Rightarrow I={{e}^{k}}\log k-\int{{{e}^{k}}\times \dfrac{1}{k}dk}+\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk} \\
\end{align}$
Now, in the integral $\int{\left( {{e}^{k}}\times \dfrac{1}{k} \right)dk}$ again using the ILATE rule for numbering we have to assume $\dfrac{1}{k}$ as function 1 and ${{e}^{k}}$ as function 2, so again applying the formula of integration by parts we get,
$\begin{align}
  & \Rightarrow I={{e}^{k}}\log k-\left[ \dfrac{1}{k}\times {{e}^{k}}-\int{{{e}^{k}}\times \left( -\dfrac{1}{{{k}^{2}}} \right)dk} \right]+\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk} \\
 & \Rightarrow I={{e}^{k}}\log k-\dfrac{{{e}^{k}}}{k}-\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk}+\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk} \\
\end{align}$
Cancelling the common terms and substituting back the value of k in terms of x we get,
$\therefore I=x\left( \log \left( \log x \right)-\dfrac{1}{\log x} \right)+c$
Here ‘c’ is the constant of integration as we are evaluating an indefinite integral.

Note: Note that we if we will start evaluating the second integral which is $\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk}$ then we will never be able to cancel the common integral terms that we cancelled above because the integral $\int{\dfrac{{{e}^{k}}}{{{\left( k \right)}^{2}}}dk}$ is never ending. It is completely on practice of questions that you will be able to understand which of the integral must be evaluated and which must be left so that it might get cancelled.