Mathematics

# Solve $\int {\left[ {\log \,\left( {logx} \right) + \dfrac{1}{{{{\left( {\log x} \right)}}}}} \right]dx}$

$I=x\log \left( \log x \right)+C$

##### SOLUTION

Consider the given integral.

$I=\left[ \int{\log \left( \log x \right)}+\dfrac{1}{\log x} \right]dx$

We know that

$\int{uvdx=u\int{vdx-\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}}}dx$

Therefore,

$I=\left[ \int{\log \left( \log x \right)}+\dfrac{1}{\log x} \right]dx$

$I=\int{\log \left( \log x \right)}x-\int{\dfrac{1}{\log x}\times \dfrac{1}{x}\times xdx}+\int{\dfrac{1}{\log x}dx}$

$I=x\log \left( \log x \right)-\int{\dfrac{1}{\log x}dx}+\int{\dfrac{1}{\log x}dx}$

$I=x\log \left( \log x \right)+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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