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Question: Evaluate \[\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} \] A.\[\log \left( {1 - \log x} \right) + c\] ...

Evaluate ∫1x.log⁡x dx\int {\dfrac{1}{{x.\log x}}{\text{ }}dx}
A.log⁡(1−log⁡x)+c\log \left( {1 - \log x} \right) + c
B.log⁡(log⁡(ex)−1)+c\log \left( {\log (ex) - 1} \right) + c
C.log⁡(log⁡x−1)+c\log \left( {\log x - 1} \right) + c
D.log⁡(log⁡x+x)+c\log \left( {\log x + x} \right) + c
E.log⁡(log⁡x)+c\log \left( {\log x} \right) + c

Explanation

Solution

Hint : We have an indefinite integral; we can solve this easily using the substitution method. We substitute log⁡x=t\log x = t and we change the independent variable to ‘t’. After applying the integration and simplifying we must keep the answer in terms of ‘x’ only.

Complete step-by-step answer :
Given, ∫1x.log⁡x dx\int {\dfrac{1}{{x.\log x}}{\text{ }}dx}
Now substitute log⁡x=t\log x = t.
Now differentiating we have
1xdx=dt\dfrac{1}{x}dx = dt
Now the given integral becomes
∫1x.log⁡x dx=∫1t dt\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} = \int {\dfrac{1}{t}{\text{ }}dt}
=∫1t dt= \int {\dfrac{1}{t}{\text{ }}dt}
We know the integration of 1t\dfrac{1}{t} with respect to ‘t’ is log⁡(t)\log (t). That is
=log⁡(t)+c= \log (t) + c
But initially we have substituted log⁡x=t\log x = t, then we have.
=log⁡(log⁡(x))+c= \log (\log (x)) + c, where ‘c’ is the integration constant.
Thus we have,
∫1x.log⁡x dx=log⁡(log⁡(x))+c\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} = \log (\log (x)) + c
Hence, the correct option is (e).
So, the correct answer is “Option E”.

Note : If we have a definite integral, we do not write the integration constant, because of the presence of upper and lower limits. Also, integration is a reverse process of differentiation, for example in above we have written that integration of 1x\dfrac{1}{x} is log⁡x\log x. The differentiation of log⁡x\log x is 1x\dfrac{1}{x}.