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

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

We have,

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

$\Rightarrow I =\int e^\theta\left(1+\tan \theta+\tan ^2 \theta\right) d \theta$, where $x=\tan \theta$

$\Rightarrow I =\int e^\theta\left(\sec ^2 \theta+\tan \theta\right) d \theta$

$\Rightarrow I =e^\theta \tan \theta+C ~~~~~~~\left[∵ \int e^x\left\{f(x)+f'(x)\right\} d x=e^x f(x)\right]$

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