${\displaystyle \int e^{{\tan }^{-1}x}}(1+x+x^2).d\left({\cot }^{-1}x\right)$  is equal to

$I=\int e^{{\tan }^{-1}x}(1+x+x^2)\frac{-1}{1+x^2}dx$ 

    $=-{\displaystyle \int e^{{\tan }^{-1}x}\left(1+\frac{x}{1+x^2}\right)dx}$

     $=-{\displaystyle \int e^{{\tan }^{-1}x}dx-{\displaystyle \int }x.\frac{e^{{\tan }^{-1}x}}{1+x^2}dx}$

     $=-{\displaystyle \int e^{{\tan }^{-1}x}dx-x}.e^{{\tan }^{-1}x}+{\displaystyle \int e^{{\tan }^{-1}x}dx+C}$

      $=-x.e^{{\tan }^{-1}x}+C$