Question

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

Answer 1

Hint: We have an indefinite integral; we can solve this easily using the substitution method. We substitute \[\log x = t\] 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, \[\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} \]
Now substitute \[\log x = t\].
Now differentiating we have
\[\dfrac{1}{x}dx = dt\]
Now the given integral becomes
 \[\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} = \int {\dfrac{1}{t}{\text{ }}dt} \]
\[ = \int {\dfrac{1}{t}{\text{ }}dt} \]
We know the integration of \[\dfrac{1}{t}\] with respect to ‘t’ is \[\log (t)\]. That is
\[ = \log (t) + c\]
But initially we have substituted \[\log x = t\], then we have.
\[ = \log (\log (x)) + c\], where ‘c’ is the integration constant.
Thus we have,
\[\int {\dfrac{1}{{x.\log x}}{\text{ }}dx} = \log (\log (x)) + c\]
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 \[\dfrac{1}{x}\] is \[\log x\]. The differentiation of \[\log x\] is \[\dfrac{1}{x}\].