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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.

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

Answers (1)

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

Solution:

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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