Mathematics

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


ANSWER

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


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
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