$$
\begin{align}
\int\frac{x^2}{\sqrt{1+x^2}}e^{\arctan(x)}\,\mathrm{d}x
&=\int\tan^2(u)\sec(u)e^u\,\mathrm{d}u\\
&=\int(\sec^3(u)-\sec(u))e^u\,\mathrm{d}u
\end{align}
$$
Now consider
$$
\begin{align}
\int\sec(u)e^u\,\mathrm{d}u
&=e^u\sec(u)-\int\sec(u)\tan(u)e^u\,\mathrm{d}u\\
&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(\sec(u)\tan^2(u)+\sec^3(u))e^u\,\mathrm{d}u\\
&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(2\sec^3(u)-\sec(u))e^u\,\mathrm{d}u\\
0&=e^u\sec(u)-e^u\sec(u)\tan(u)+\int(2\sec^3(u)-2\sec(u))e^u\,\mathrm{d}u\\