ยากเหมือนกันนะครับ มองอยู่นานทีเดียว แต่ผมไม่ integrated mania ทำได้ไหมครับพี่ passer-by
Let $u=\arctan x$. We can see that \[ \int \frac{x^2e^{\arctan x}}{\sqrt{1+x^2}} dx = \int e^u\tan^2 u \sec u du \]
Consider
\[ \int e^u\tan^2 u \sec u \; du = \int e^u\sec^3u \; du - \int e^u\sec u \; du \]
Notice that ${\displaystyle \int e^u \sec ^3 u du = \int e^u\sec u \; d(\tan u) }$. Using the integration by parts, we obtain \[ \begin{array}{ccc}{\displaystyle \int e^u \sec ^3 u \; du } &=& {\displaystyle e^u \sec u \; \tan u - \int \tan u (e^u \sec u \; \tan u + e^u\sec u) \; du } \\
&=& {\displaystyle e^u \sec u \; \tan u - \int e^u \tan ^2 u \sec u du - \int e^u\tan u\sec u \; du } \end{array} \].
Using the integration by parts again, ${\displaystyle \int e^u\tan u\sec u \; du = e^u\sec u - \int e^u\sec u \; du }$
Hence, \[\begin{array}{ccl}{\displaystyle \int e^u\tan^2 u \sec u \; du } &=& {\displaystyle e^u \sec u \; \tan u - \int e^u \tan ^2 u \sec u du - \int e^u\tan u\sec u \; du - \int e^u\sec u \; du} \\
{\displaystyle 2 \int e^u\tan^2 u \sec u } &=& {\displaystyle e^u \sec u \; \tan u -(e^u\sec u - \int e^u\sec u du) - \int e^u\sec u \; du} \\
{\displaystyle \int e^u\tan^2 u \sec u } & = &{\displaystyle \frac{1}{2} \left[ e^u \sec u \; \tan u - e^u\sec u \right] + C } \end{array}\]
Replace $u=\arctan x $, so we are done.