Who's up to the challenge? 5

Let the integral be $I$ , then we have:

$\begin{aligned} I & = \int_0^1 x^4 e^x \sin x \space dx \\ & = \Im \int_0^1 x^4 e^x e^{ix} \space dx \\ & = \Im \int_0^1 x^4 e^{(1+i)x} \quad \quad \small \color{#3D99F6}{\text{By integration by parts.}} \\ & = \Im \left[ \frac{x^4e^{(1+i)x}}{1+i} - \frac{4x^3e^{(1+i)x}}{(1+i)^2} + \frac{12x^2e^{(1+i)x}}{(1+i)^3} - \frac{24xe^{(1+i)x}}{(1+i)^4} + \frac{24e^{(1+i)x}}{(1+i)^5} \right]_0^1 \\ & = \Im \left( \frac{e^{1+i}}{1+i} - \frac{4e^{1+i}}{(1+i)^2} + \frac{12e^{1+i}}{(1+i)^3} - \frac{24e^{1+i}}{(1+i)^4} + \frac{24e^{1+i}}{(1+i)^5} - \frac{24}{(1+i)^5} \right) \\ & = \Im \left(e^{1+i}\left[\frac{1}{2}(1-i) + 2i - 3(1+i) + 6 - 3(1-i)\right] + 3(1-i) \right) \\ & = \Im \left( \frac{e(\cos 1 + i \sin 1)(1+3i)}{2} + 3(1-i) \right) \\ & = \frac{e}{2}(\sin 1 + 3 \cos 1 ) - 3 \end{aligned}$

$\Rightarrow A + B + C + D + E + F = 1+2+1+3+1+3 = \boxed{11}$