Is this product and sum of integrals divisible by the polynomial $x^4$? Problem:

Let $A(t)$, $B(t)$, $C(t)$ and $D(t)$ be any polynomials with real
  coefficients. Show that,

$F(x)$=

is divisible by $x^4$.


Proff:
Before all, thanks LutzL for your clear explanation, here I put all the pieces together, to have a better understanding and to ask if I need to put some extra step in the proof explaining more, (I believe is fine, but I never sure). 
I put this symbol [] to inform that all was written by LutzL [“from LutzL comments and answer”].

F(X)=



This formula [‘has the form of a scalar product of two cross products
  or the Cauchy-Binet formula which says that
  (a×c,b×d)=(a,b)·(c,d)−(a,d)·(b,c)’]
And the right, part [ ‘on the right you see the form of your given
  term, on the left the components are (modulo signs)



giving the scalar product,



and we get;



Now one has to explore the cancellation of lower-order terms. The integrand is a multiple of (s−t)^2  ds dt, which already accounts for the factor x^4. ]



And then substitute s=xu , t=xv to get an equal form of the integral
  of

**

The integrand is still a polynomial in u, v, x so that integration gives a polynomial in x as result. As for having two different integration variables, allows writing everything as one integral over a square.


In conclusion, from ** we can see that F(X)= (X^4 ) *  H(x) , where H(x) is the integral over the square. Therefore, F(x) is divisible by x^4.
Question: I need to explain, something else or is already a valid proof?
Again: Thanks, LutzL.
 A: I will write $A(t) = a_0 + a_1t + O(t^2)$, and the same for other polynomials. Then
\begin{align}
\int_0^x A(t)B(t) dt & = \int_0^x \left(a_0b_0 + (a_1b_0 + b_0b_1)t + O(t^2)\right) dt \\
& = a_0b_0x + \frac 12(a_1b_0 + a_0b_1)x^2 + O(x^3)
\end{align}
Therefore,
\begin{align}
& \left(\int_0^x A(t)B(t)dt\right)\left(\int_0^x C(t)D(t)dt\right) \\
& \quad = a_0b_0c_0d_0x^2 + \frac 12(a_1b_0c_0d_0 + a_0b_1c_0d_0 + a_0b_0c_1d_0 + a_0b_0c_0d_1)x^3 + O(x^4).
\end{align}
It is obvious that permuting $(A, B, C, D)$ does not change the part with degrees lower than $4$.
A: One can write the double of the term as
$$
\int_0^x\int_0^x (A(s)C(t)-A(t)C(s))·(B(s)D(t)-B(t)D(s))\,dsdt
$$
Now one has to explore the cancellation of lower-order terms. The integrand is a multiple of $(s-t)^2dsdt$, which already accounts for the factor $x^4$. More concretely, set 
$$
A(s)C(t)-A(t)C(s)=(s-t)P(s,t),\\ 
B(s)D(t)-B(t)D(s)=(s-t)Q(s,t)
$$
and then substitute $s=xu$, $t=xv$ to get an equal form of the integral of
$$
x^4·\int_0^1\int_0^1 (u-v)^2·P(ux,vx)·Q(ux,vx)\,dudv
$$
A: One way is to start by proving it for one term monic polynomials. Let $A(t)=t^a$ and so on.  Then $\int_0^x t^a dt=\frac {t^{a+1}}{a+1}$ and the whole expression becomes $$\left(\frac 1{(a+b+1)(c+d+1)}-\frac 1{(a+d+1)(b+c+1)}\right)x^{a+b+c+d+2}$$  If at least two of $a,b,c,d$ are positive, the exponent on $x$ is at least $4$.  If three of them are zero, the coefficient is zero. In either case we have the desired result.  Then show you can multiply any polynomial by a constant and it still works.  Now assume it has been proven for the combinations $a_1(t),B(t),C(t),(D(t)$ and $a_2(t),B(t),C(t),(D(t)$.  Show it still holds for $(a_1+a_2)(t),B(t),C(t),(D(t)$  by the linearity of the integral.  Now argue you can build up integrals of any complexity required by steps like this.
