How do you integrate $ \left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx $ ?

Hint: To solve the above question we can apply a substitution method. If we are given a equation like $ \int{f\left( g\left( x \right) \right)g'\left( x \right)dx} $ to integrate where $ g'\left( x \right) $ is differentiation of $ g\left( x \right) $ with respect to x , we can substitute $ g\left( x \right) $ with t so the value of $ g'\left( x \right)dx $ will equal to $ dt $ so equation will change to $ \int{f\left( t \right)dt} $ . We can apply this method to solve the above question.

Complete step by step answer:
In the given question we have integrate $ \left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx $
We can see there are 2 terms present in the equation $ \ln x $ and $ \dfrac{1}{x} $ and we know that derivative of $ \ln x $ is $ \dfrac{1}{x} $ so it is a good idea to take $ \ln x $ as t
If $ t=\ln x $ , then we can write
 $ dt=\dfrac{1}{x}dx $
Now replacing $ \ln x $ with t and $ \dfrac{1}{x}dx $ with $ dt $ we can write
 $ \int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx=\int{tdt}} $
We know that the value of $ \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c $ where n is not equal to -1 and c is a constant.
When we put x=1 we get $ \int{xdx=\dfrac{{{x}^{2}}}{2}}+c $
We can write $ \int{tdt=\dfrac{{{t}^{2}}}{2}+c} $

Now we know that $ t=\ln x $ we can replace t with $ \ln x $

 $ \int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx=\int{\dfrac{{{\left( \ln x \right)}^{2}}}{2}}}+c $ where c is any constant.
This is one way we can solve the question.

Note:
 Another method to solve the integration is by integration by parts . In integration by parts we write $ \int{f\left( x \right)g\left( x \right)dx=}f\left( x \right)\int{g\left( x \right)dx-\int{f'\left( x \right)\left( \int{g\left( x \right)dx} \right)dx+c}} $ where $ f'\left( x \right) $ is derivative of $ f\left( x \right) $ with respect to x. So we can write
 $ \Rightarrow \int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx=}\ln x\int{\dfrac{1}{x}dx-\int{\dfrac{d\left( \ln x \right)}{dx}.\left( \int{\dfrac{1}{x}dx} \right)}dx}+c $
We know that the integration of $ \dfrac{1}{x} $ is $ \ln x $ and derivative of $ \ln x $ is $ \dfrac{1}{x} $
 $ \Rightarrow \int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx=}\ln x.\ln x-\int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx}+c $
Now if we take the variable I as $ \int{\ln \left( x \right).\left( \dfrac{1}{x} \right)dx} $ and replace in our equation we get
 $ I={{\left( \ln x \right)}^{2}}-I+c $
By further solving
 $ \Rightarrow 2I={{\left( \ln x \right)}^{2}}+c $
 $ \Rightarrow I=\dfrac{{{\left( \ln x \right)}^{2}}}{2}+{{c}_{1}} $
So $ \int{\left( \ln \left( x \right).\left( \dfrac{1}{x} \right) \right)dx=\int{\dfrac{{{\left( \ln x \right)}^{2}}}{2}}}+{{c}_{1}} $ where $ {{c}_{1}} $ is any constant and $ {{c}_{1}}=\dfrac{c}{2} $