∫ e^tan^-1x(1+x+x^2)d(cot^-1x)=

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

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

$=-\int e^{\tan^{-1}x}dx-\{e^{\tan^{-1}x}.x-\int e^{\tan^{-1}x}.1\,dx\}+c$

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