Integrating $\int^1_0 \frac{x^2e^{\arctan x}}{\sqrt{x^2+1}}\,dx$ This is a very hard integral that I am trying to solve. I’ve tried many substitutions, integration by parts, but I cannot evaluate this. Are there any other approaches I can take to solve this
$$\int^1_0 \dfrac{x^2e^{\arctan x}}{\sqrt{x^2+1}}\,dx$$

Let $\displaystyle{I = \int^1_0 \frac{x^2 e^{\tan^{-1} x}}{\sqrt{x^2 + 1}} \, dx}.$
Let $x = \tan u, dx = \sec^2 u \, du.$ Limits: $ x = 1, u = \pi/4$ and $x = 0, u = 0.$ So
\begin{align*}I &= \int^{\frac{\pi}{4}}_0 \frac{\tan^2 u e^u}{\sqrt{\tan^2 u + 1}} \cdot \sec^2 \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u \tan^2 u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u (\sec^2 u - 1) \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 e^u \sec^3 u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &= \int^{\frac{\pi}{4}}_0 \sec^2 u \cdot e^u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &= \Big{[}\tan u \cdot e^u \sec u \Big{]}^{\frac{\pi}{4}}_0 - \int^{\frac{\pi}{4}}_0 \tan u (e^u \sec u + e^u \sec u \tan u) \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du \quad \mbox{(by parts)}\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_\alpha e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \tan^2 u \sec u \, du -\int^{\frac{\pi}{4}}_0 e^u \sec u \, du\end{align*}
Since the second integral here is the integral we started with, we have
\begin{align*}\Rightarrow 2I &= \sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du - \int^{\frac{\pi}{4}}_0 e^u \sec u \, du\\ &=\sqrt{2} e^{\frac{\pi}{4}} - \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du - \Big{[}e^u \sec u \Big{]}^{\frac{\pi}{4}}_0 + \int^{\frac{\pi}{4}}_0 e^u \tan u \sec u \, du \quad \mbox{(by parts)}\\&=\sqrt{2} e^{\frac{\pi}{4}} - \sqrt{2} e^{\frac{\pi}{4}} + \frac{1}{2}\\ &= \frac{1}{2}\end{align*}
