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

 

Putting ${\tan }^{-1}x=u,\text{ we have }\frac{\mathrm{d}x}{1+x^2}=\mathrm{d}u$

$\begin{array}{l}\int e^{{\tan }^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)\mathrm{d}x=\int e^u\left(1+\tan u+{\tan }^{2}u\right)\mathrm{d}u\\ =\int e^u\left({\sec }^{2}u+\tan u\right)\mathrm{d}u\\ =\tan u{e}^u+C=x{e}^{{\tan }^{-1}x}+C.\end{array}$

$\text{ (using }\int e^x\left(f\left(x\right)+f^{\mathrm{\prime }}\left(x\right)dx=e^{x}f\left(x\right)+C\right.\text{ ) }$