Let $$F : [3, 5] \rightarrow R$$ be a twice differentiable function on $$(3, 5)$$ such that $$F(x) = e^{-x} \int_3^x (3t^2 + 2t + 4F'(t)) \, dt$$. If $$F'(4) = \frac{\alpha e^\beta - 224}{(e^\beta - 4)^2}$$, then $$\alpha + \beta$$ is equal to _________.

We have a function $$F : [3,5]\rightarrow \mathbb R$$ which is twice differentiable on $$(3,5)$$ and is defined by

$$F(x)=e^{-x}\int_{3}^{x}\bigl(3t^{2}+2t+4F'(t)\bigr)\,dt.$$

Write the integral inside as $$G(x)=\displaystyle\int_{3}^{x}\bigl(3t^{2}+2t+4F'(t)\bigr)\,dt,$$ so that the given relation becomes $$F(x)=e^{-x}G(x).$$

Differentiate both sides with respect to $$x$$. Using the product rule for $$e^{-x}G(x)$$ we obtain

$$F'(x)=\bigl(-e^{-x}\bigr)G(x)+e^{-x}G'(x)=e^{-x}\bigl(G'(x)-G(x)\bigr).$$

But $$G'(x)$$ is simply the integrand evaluated at $$t=x$$, therefore

$$G'(x)=3x^{2}+2x+4F'(x).$$

Substituting this value of $$G'(x)$$ in the derivative we just found gives

$$F'(x)=e^{-x}\bigl(3x^{2}+2x+4F'(x)-G(x)\bigr).$$

Because $$G(x)=e^{x}F(x)$$ (directly from $$F(x)=e^{-x}G(x)$$), the above becomes

$$F'(x)=e^{-x}\bigl(3x^{2}+2x+4F'(x)-e^{x}F(x)\bigr).$$

Multiplying through by $$e^{x}$$ yields

$$e^{x}F'(x)=3x^{2}+2x+4F'(x)-e^{x}F(x).$$

Gather the $$F'(x)$$ terms on the left:

$$(e^{x}-4)F'(x)=3x^{2}+2x-e^{x}F(x).$$

Divide by $$e^{x}-4\;(\neq 0\text{ on }(3,5))$$ to obtain a linear differential equation in $$F$$:

$$F'(x)+\frac{e^{x}}{e^{x}-4}\,F(x)=\frac{3x^{2}+2x}{e^{x}-4}.$$

The standard form $$y'+P(x)y=Q(x)$$ suggests an integrating factor. We first state the formula: for $$y'+P(x)y=Q(x)$$, the integrating factor is $$\mu(x)=e^{\int P(x)\,dx}.$$ Here $$P(x)=\dfrac{e^{x}}{e^{x}-4}.$$

Compute the integral:

$$\int\frac{e^{x}}{e^{x}-4}\,dx.$$ Let $$u=e^{x}-4\;\Rightarrow\;du=e^{x}\,dx.$$ Hence the integral becomes $$\int\frac{1}{u}\,du=\ln|u|.$$

So $$\mu(x)=e^{\ln|e^{x}-4|}=e^{x}-4.$$ (For $$x\in(3,5),\;e^{x}-4>0$$, so absolute value is unnecessary.)

Multiply the differential equation by this integrating factor:

$$(e^{x}-4)F'(x)+e^{x}F(x)=3x^{2}+2x.$$

The left side is the derivative of $$(e^{x}-4)F(x)$$, because

$$\frac{d}{dx}\bigl((e^{x}-4)F(x)\bigr)=e^{x}F(x)+(e^{x}-4)F'(x).$$

Therefore

$$\frac{d}{dx}\bigl((e^{x}-4)F(x)\bigr)=3x^{2}+2x.$$

Integrate both sides from $$3$$ to $$x$$:

$$(e^{x}-4)F(x)-(e^{3}-4)F(3)=\int_{3}^{x}(3t^{2}+2t)\,dt.$$

Compute the definite integral on the right:

$$\int(3t^{2}+2t)\,dt=t^{3}+t^{2},$$ so

$$\int_{3}^{x}(3t^{2}+2t)\,dt=(x^{3}+x^{2})-(3^{3}+3^{2})=x^{3}+x^{2}-36.$$

Next, evaluate $$F(3)$$ from the original definition. Because the upper and lower limits of the integral coincide at $$x=3$$, we have $$\int_{3}^{3}\cdots=0,$$ hence $$F(3)=0.$$

Putting $$F(3)=0$$ into the integrated equation gives

$$(e^{x}-4)F(x)=x^{3}+x^{2}-36.$$

Thus the explicit expression for $$F(x)$$ is

$$F(x)=\frac{x^{3}+x^{2}-36}{e^{x}-4}.$$

We now need $$F'(4).$$ From an earlier step we already have an algebraic relation for $$F'(x$$):

$$F'(x)=\frac{3x^{2}+2x-e^{x}F(x)}{e^{x}-4}.$$

First evaluate $$F(4)$$ using the formula for $$F(x):$$

$$F(4)=\frac{4^{3}+4^{2}-36}{e^{4}-4}=\frac{64+16-36}{e^{4}-4}=\frac{44}{e^{4}-4}.$$

Substitute $$x=4$$ and $$F(4)$$ into the expression for $$F'(x)$$:

$$F'(4)=\frac{3(4)^{2}+2(4)-e^{4}F(4)}{e^{4}-4}.$$

Simplify the numerator step by step:

$$3(4)^{2}=3\cdot16=48,$$ $$2(4)=8,$$ so $$3(4)^{2}+2(4)=48+8=56.$$

Also, $$e^{4}F(4)=e^{4}\cdot\frac{44}{e^{4}-4}=\frac{44e^{4}}{e^{4}-4}.$$

Hence the numerator becomes

$$56-\frac{44e^{4}}{e^{4}-4}=\frac{56(e^{4}-4)-44e^{4}}{e^{4}-4}.$$

Expand and combine like terms:

$$56e^{4}-224-44e^{4}=12e^{4}-224.$$

Therefore

$$F'(4)=\frac{12e^{4}-224}{(e^{4}-4)^2}.$$

This matches the required form $$F'(4)=\dfrac{\alpha e^\beta-224}{(e^\beta-4)^2},$$ from which we read $$\alpha=12,\qquad\beta=4.$$

Adding these two constants gives $$\alpha+\beta=12+4=16.$$

So, the answer is $$16$$.