$\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x$
$\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x$
$\left(\frac{x}{2}\right) e^{\tan ^{-1} x}+c$
$x e^{\tan ^{-1} x}+c$
$\left(\frac{1}{2}\right) e^{\tan ^{-1} x}+c$
$e^{\tan ^{-1} x}+c$
Solution
Let $I=\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x$
$\begin{aligned} I &=\int e^{\tan ^{-1} x}\left(\frac{1+x^{2}+x}{1+x^{2}}\right) d x \\ \text { Put } \tan ^{-1} x &=t \Rightarrow x=\tan t \text { and } \frac{1}{1+x^{2}} d x=d t \\ \therefore I &=\int e^{t}\left(1+\tan ^{2} t+\tan t\right) d t \Rightarrow \int e^{t}\left(\tan t+\sec ^{2} t\right) d t \\ &=e^{t} \tan t+c=x e^{\tan ^{-1} x}+c \end{aligned}$