Mathematics

# Integration of $\cfrac { 1 }{ 1+{ \left( \log _{ e }{ x } \right) }^{ 2 } }$ with respect to $\log _{ e }{ x }$ is

$\tan ^{ -1 }{ \left( \log _{ e }{ x } \right) }$

##### SOLUTION
$I=\displaystyle\int{\dfrac{1}{1+{\left(\log_{e}{x}\right)}^{2}}d\left(\log_{e}{x}\right)}$

$=\displaystyle\int{\dfrac{dx}{x\left(1+{\left(\log_{e}{x}\right)}^{2}\right)}}$

Let $t=\log_{e}{x}\Rightarrow\,dt=\dfrac{1}{x}dx$

$=\displaystyle\int{\dfrac{dt}{1+{t}^{2}}}$

$={\tan}^{-1}{t}+c$

$={\tan}^{-1}{\log_{e}{x}}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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On the basis of above information, answer the following questions :

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1 Verified Answer | Published on 17th 09, 2020