Mathematics

The value of $$\int _{ 1/e }^{ \tan { x }  }{ \dfrac { t }{ 1+{ t }^{ 2 } } dt+ } \int _{ 1/e }^{ \cot { x }  }{ \dfrac { t }{ t\left( 1+{ t }^{ 2 } \right)  } dt } $$, where $$x\in \left( \pi /6,\ \pi /3 \right) $$, is equal to :


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$$0$$


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Single Correct Medium Published on 17th 09, 2020
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