The value of _e^4^e^2(1)/(x)(e^(( _ex)^2+1)^-1e^(( _ex)^2+1)^-1+e^((6- _ex)^2+1)^-1)dx is

Question

The value of ${\int }_{{e}^{4}}^{{e}^{2}}\frac{1}{x}(\frac{{e}^{{({({\log }_{e}x)}^{2}+1)}^{-1}}}{{e}^{{({({\log }_{e}x)}^{2}+1)}^{-1}}+{e}^{{({(6-{\log }_{e}x)}^{2}+1)}^{-1}}})dx$ is

Accepted Answer

$\begin{gathered} \ln x=t\Rightarrow \frac{dx}{x}=dt \ I={\int }_{4}^{2}\frac{{e}^{\frac{1}{1+{t}^{2}}}}{{e}^{\frac{1}{1+{t}^{2}}}+{e}^{\frac{1}{1+(6-t{)}^{2}}}}dt \ I={\int }_{4}^{2}\frac{\frac{1}{{e}^{1+(6-t{)}^{2}}}}{\frac{1}{{e}^{1+(6-t{)}^{2}}}+{e}^{\frac{1}{1+{t}^{2}}}}dt \ 2I={\int }_{4}^{2}dt=(t{)}_{4}^{2}=4-2=2 \ I=1 \end{gathered}$

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