u = \arctan x\\
du = \frac{1}{1+x^2} dx\\
dx = (1+x^2)du\\
x = \tan u\\
\displaystyle \frac{xe^{\arctan x}}{(1+x^2)^{3/2}}dx = \frac{\tan u \cdot e^u}{(1+(\tan u)^2)^{3/2}}(1+(\tan u)^2)du = \frac{\tan u \cdot e^u}{\sqrt{1+(\tan u)^2}}du = \frac{\tan u \cdot e^u}{\frac{1}{\cos u}}du\\ = \sin u \cdot e^u \, du