TITLE: Evaluate $\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$
QUESTION [7 upvotes]: $$\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$$
My try:
I tried substituting $x= \tan\theta$ and got,
$\displaystyle\int e^\theta \tan^2\theta \sec\theta\ \mathrm{d}\theta$ and cannot move forward. so I thought of finding two integrals($I$ and $J$)one of which is required and the other is taken such that $I+J, I-J$ are easy to find.but couldn't find such $J$ $\left(I=\displaystyle\int\frac{x^2e^{\arctan x}}{1+x^2}\ \mathrm{d}x\right)$
Also, I don't think I could make use of definite integral properties.
WolframAlpha result for the integral is
$\frac{(x-1)\sqrt{1+x^2}e^{arctanx}}{2}+C$
I also want to know how you got the idea for the solution.

REPLY [3 votes]: Substitute $x=\tan t$
\begin{align}
I= \int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ {d}x
&= \int_{-\pi/4}^{\pi/4} e^{t}\tan^2 t\sec t\>dt\\
&=\int_{-\pi/4}^{\pi/4} e^{t}\sec t \> d(\tan t-1)
- \int_{-\pi/4}^{\pi/4} e^{t}\sec t \> dt\\
&= e^{t}\sec t (\tan t - 1) \bigg|_{-\pi/4}^{\pi/4}- I=\sqrt2e^{-\frac\pi4}
\end{align}