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

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

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

$= \int {{e^{{{\tan }^{ - 1}}x}}} dx + \int {\frac{{x{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}} dx$
$I = {I_1} + {I_2}$

……..(i)
Now, ${I_2} = \int {\frac{{x{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}} dx$

Let's put ${\tan ^{ - 1}}x = t \Rightarrow x = \tan t$
$\Rightarrow$ $\frac{1}{{1 + {x^2}}}dx = dt$

therefore,$I = \int {\mathop {\tan t \cdot }\limits_{\rm{I}} } \,\mathop {{e^t}}\limits_{{\rm{II}}} dt$

$= \tan t \cdot {e^t} - \int {{{\sec }^2}} t \cdot {e^t}dt + C$

${I_2} = \tan t \cdot {e^t} - \int {\left( {1 + {x^2}} \right)} \frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx + C$

${I_2} = \tan t \cdot {e^t} - \int {{e^{{{\tan }^{ - 1}}x}}} dx + C$

therefore,$I = \int {{e^{{{\tan }^{ - 1}}x}}} dx + \tan t \cdot {e^t} - \int {{e^{{{\tan }^{ - 1}}x}}} dx + C$

$= \tan t \cdot {e^t} + C$

$= x{e^{{{\tan }^{ - 1}}x}} + C$