f(x) = e^(3·x)·(2·x^3 + 0.5·x^2 + 1)
F(x) = e^(3·x)·(a·x^3 + b·x^2 + c·x + d)
F'(x) = e^(3·x)·((3·a)·x^3 + (3·a + 3·b)·x^2 + (2·b + 3·c)·x + (c + 3·d))
3·a = 2 --> a = 2/3
3·(2/3) + 3·b = 0.5 --> b = -1/2
2·(-0.5) + 3·c = 0 --> c = 1/3
(1/3) + 3·d = 1 --> d = 2/9
F(x) = e^(3·x)·(2/3·x^3 - 1/2·x^2 + 1/3·x + 2/9)