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

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

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

$\int\left(\frac{1+x+x^2}{1+x^2}\right)e^{\tan^{-1}x}dx$ $=\int\left(1+\frac{x}{1+x^2}\right)e^{\tan^{-1}x}dx$ this is of the form $\int e(f(x)+f'(x))dx$ as $\frac{dxe^{\tan^{-1}x}}{dx}=e^{\tan^{-1}x}+\frac{xe^{\tan^{-1}x}}{1+x^2}$ so $\int\left(1+\frac{x}{1+x^2}\right)e^{\tan^{-1}x}dx=xe^{\tan^{-1}x}+c$