\[f(x) = x\sin{2x};\ \ \ x = \pi.\]
\[1)\ f(\pi) = \pi \bullet \sin{2\pi} = \pi \bullet 0 = 0;\]
\[2)\ f^{'}(x) = \sin{2x} + x \bullet 2\cos{2x};\]
\[f^{'}(\pi) = \sin{2\pi} + 2\pi\cos{2\pi} = 2\pi;\]
\[3)\ f^{'}(x) + f(x) + 2 = 2\pi + 2 =\]
\[= 2(\pi + 1);\]
\[Ответ:\ \ 2(\pi + 1).\]