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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 10 Maths Textbook Solution.

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Answer: \log x\{\log (\log x)-1\}+c

Hint:Take \log x=t \text { So, } \frac{1}{x} d x=d t

Then,\int \frac{\log (\log x)}{x} d x=\int \log t d t=\int 1 \cdot \log t d t

Consider the 1st function as log t and 2nd function

Given:  I=\int \frac{\log (\log x)}{x} d x


Let \log x=t \text { So, } \frac{1}{x} d x=d t

\begin{aligned} &I=\int 1 \cdot \log t d t \\ &=\log t \int 1 d t-\int\left(\frac{d \log t}{d t} \cdot \int 1 d t\right) d t \end{aligned}

By integrating w.r.t ‘t’

=t \log t-\int \frac{1}{t} \cdot t d t

Again by integrating the second term we get

=t \log t-t+c

Replacing t with logx

=\log x \cdot \log (\log x)-\log x+c

By taking log x as common

=\log x\{\log (\log x)-1\}+c

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