Question

How do you integrate \[\dfrac{\ln \left( \ln x \right)}{x}dx\]?

Answer 1

Hint: From the question given, we have been asked to integrate \[\dfrac{\ln \left( \ln x \right)}{x}dx\].We can solve the given question by using the basic formulae of integration. By using the basic formulae of integration, we can solve the given question very easily. We have to use the basic formula of integration that is integration by parts formula to solve the given question. Integration by parts formula is given as: \[\int{udv=uv-\int{vdu}}\]

Complete step by step answer:
From the question, we have been given that \[\dfrac{\ln \left( \ln x \right)}{x}dx\]
First of all, let us assume that \[z=\ln x\]
Then, by differentiating it, we get \[\Rightarrow dz=\dfrac{dx}{x}\]
Now, we can write the given question as \[\Rightarrow \int{\dfrac{\ln \left( \ln x \right)}{x}dx=\int{\ln zdz}}\]
Integration by parts formula is given as: \[\int{udv=uv-\int{vdu}}\]
Now, we have to use the integration by parts formula to the above obtained equation to get the final solution for the given question.
By comparing the terms, we get \[\Rightarrow u=\ln z,v=z\]
Therefore \[\Rightarrow \int{\ln zdz}=\ln z\times z-\int{z\dfrac{dz}{z}}\]
Simplify further to get the final answer \[\Rightarrow \int{\ln zdz}=z\ln z-z\]
Hence, \[\Rightarrow \int{\dfrac{\ln \left( \ln x \right)}{x}dx=\ln x\ln \left( \ln x \right)-\ln x+c}\]
Hence, we got the final answer for the given question.

Note:
We should be well known about the basic formulae of integration and also we should be well aware of the applying the basic formulae of integration. Also, we should be well known about the usage of the basic formulae for the given question. Also, we should be very careful while doing the calculation part as integral calculations are somewhat difficult to do. Also, we should be very careful while applying the integration by parts formula. Sometimes it will be helpful to remember the answers of questions of this type as formulae because they will help us in further complex calculations.