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

$\int e^{{\tan }^{-1}x}\left[\frac{{\displaystyle 1+x+x^2}}{{\displaystyle 1+x^2}}\right]dx$

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

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

$=\int e^{g\left(x\right)} \left[g\text{'}\left(x\right)f\left(x\right)+f\text{'}\left(x\right)\right]dx$

$where g\left(x\right)={\tan }^{-1}x, f\left(x\right)=x$

$$\begin{gathered} =e^{g\left(x\right)}f\left(x\right)+C \\ =e^{{\tan }^{-1}x}\left(x\right)+C \end{gathered}$$