Question

Integrate $\int{\dfrac{\log (\log x)}{x}}dx$

Answer 1

To integrate the given equation we need to consider the value of $\log x$ as $t$. So that it will become easy to integrate. After integrating the equation, replace it with $t$ $\log x$. This will give the expression for integration of $\int{\dfrac{\log (\log x)}{x}}dx$. We use the formula $\int{\log x=x\left( \log x-1 \right)}+C$ to solve the $\log x$ functions.

Complete step-by-step solution:
Given expression is $\int {\dfrac{{\log (\log x)}}{x}} dx$
Let us consider variable $y$ equal to the given expression
$y = \int {\dfrac{{\log (\log x)}}{x}} dx$
Consider $t = \log x$
We also have
$dt = \dfrac{1}{x}dx$
After that substitute the values in the given equation.
$y = \int {\dfrac{{\log (\log x)}}{x}} dx \\\ \Rightarrow \int {\dfrac{{\log (\log x)}}{x}} dx = \int {\log (t) \times \dfrac{{dx}}{x}} \\\ \Rightarrow \int {\dfrac{{\log (\log x)}}{x}} dx = \int {\log (t) \times dt}$
Solving the above term y using by arts we get:

\Rightarrow Let\,y = \int {\log \left( t \right)} dt \\\ \Rightarrow y = \int {\log (t) \times 1\,dt} \\\ \Rightarrow {1^{st}}\,function = f(x) = \log (t) \\\ \Rightarrow {2^{nd}}\,function = g(x) = 1 \\\ \Rightarrow y = \log (t)\int {1dt} - \int {\left[ {\left( {\dfrac{{d\left[ {\log (t)} \right]}}{{dt}}} \right)\int {1dt} } \right]dt} \\\ \Rightarrow y = [\log (t)]t - \int {\dfrac{1}{t} \times t} dt $$ $$ \Rightarrow y = [\log (t)]t - \int 1 dt = t\log (t) - t + c$$ Now replace $t$ with $\log x$ $\Rightarrow y = \log x\left( {\log \left( {\log x} \right) - 1} \right) + C \\\ \Rightarrow y = \log x\log \left( {\log x} \right) - \log x + C $ **Additional Information:** The formula for integration of two terms is given by $\int {uvdx = u\int {vdx - \int {u' \left(\int {vdx}\right) }dx } } $ To do integration of two or more combined equations we use I LATE rule where I stands for Inverse functions like inverse trigonometric functions L stands for Logarithm functions A stands for algebraic functions T stands for Trigonometric functions E stands for exponential functions. **Generally, we use the order of ILATE to take the value of $u$. For the exponential functions the function remains the same even after integrating it. For logarithmic functions the differentiation gives the expression $$\dfrac{1}{x}$$** **Note:** The integration of an expression is considered as the inverse of the differentiation process. The resultant of integration after differentiation is considered as the same expression. The differentiation will decrease the value and integration will increase the value of the expression. The integration of a given equation is considered as an area under the given curve equation when drawn on a graph. It can be found from various limits.