Question

Integrate the following function-
$\smallint \dfrac{{\ln \left( {\ln x} \right)}}{{x\ln x}}dx\left( {x > 0} \right)$

Answer 1

Hint: This problem requires the various formulas and methods that are used in indefinite integrals. We will first use the substitution method where we will assume lnx as a suitable variable(t), and then compute the value of dx in terms of dt. Then we can integrate the simplified form using the rules for integration.

Complete step-by-step answer:
We need to solve the integral-
$\smallint \dfrac{{\ln \left( {\ln x} \right)}}{{x\ln x}}dx\left( {x > 0} \right)$
This cannot be integrated directly, so we will try to substitute a value to simplify it. Let us assume that-
$\ln x = t$ ...(1)
We will now differentiate both the sides with respect to x and t respectively, to find dx in terms of dt. This can be done using the formula-
$\dfrac{{d\left( {\ln x} \right)}}{{dx}} = \dfrac{1}{x}$
So, we can write that-
$\ln x = t$
$\dfrac{1}{x}dx = dt$ ...(2)
We will now substitute these values in equations (1) and (2) in the integral as-
$\smallint \dfrac{{\ln t}}{t}dt$
We can see that this form can still not be integrated directly, so we will substitute a value for t to further simplify our integral as-
$\ln t = z$
Differentiating wrt t and z,
$\dfrac{1}{t}dt = dz$
We will further substitute these values of t and dt in terms of z in the integral as-
$\smallint zdz$
This integral can be easily solve using the formula-
$\smallint {x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
So we can integrate the function as-
$\begin{aligned} &= \dfrac{{{z^{1 + 1}}}}{{1 + 1}} + C \\\ &= \dfrac{{{z^2}}}{2} + C \\\ &We;\;know\;that\;z = \ln t, \\\ &= \dfrac{{{{\left( {\ln t} \right)}^2}}}{2} + C \\\ &Also;\;t = \ln x, \\\ & = \dfrac{{{{\left( {\ln \left( {\ln x} \right)} \right)}^2}}}{2} + C \\\ \end{aligned}$

This is the required value of the integral.

Note: The most common mistake made by a large number of students in any problem related to indefinite integrals is that they forget to add the constant of integration(C) at the end of the integral. Also, they often leave the answer in terms of the assumed variable(z), and don’t back substitute their values to write the answer in terms of the initial variable, which is x.