Integral functional equation. $f(x)= \Big(\int _1^x g_1(t)g_2(t)dt\Big)\Big(\int _1^x g_3(t)g_4(t)dt\Big)-\Big(\int _1^x g_1(t)g_3(t)dt\Big)\Big(\int _1^x g_2(t)g_4(t)dt\Big) $ $\forall$  $x$ $\in R$
Where $g_1(x),g_2(x),g_3(x)$ and $g_4(x)$ are non constant continuous and differentiable functions and $f(x)$ is a polynomail of degree $4$ with leading coefficent $1$.
ATTEMPT: Let $f(x)=x^4+bx^3+cx^2+dx+e.$
Then by Leibntiz's rule, $f'(x)=\Big( g_1(x)g_2(x)\Big) \Big(\int _1^x g_3(t)g_4(t)dt\Big)+\Big( g_3(x)g_4(x)\Big) \Big(\int _1^x g_1(t)g_2(t)dt\Big)-\Big( g_1(x)g_3(x)\Big) \Big(\int _1^x g_2(t)g_4(t)dt\Big)-\Big( g_2(x)g_4(x)\Big) \Big(\int _1^x g_1(t)g_3(t)dt\Big)$
However where do I employ the given conditions to get the answer?
Should I consider the integrals of $g_1(x),g_2(x),g_3(x)$ and $g_4(x)$ as $G_1(x),G_2(x),G_3(x)$ and $G_4(x)$ and try equating the coefficients of$x$?

Maybe there is a better method (well, there has to be); this one is very tedious but it works:
1) Evaluate $f(1)=0$.
2) Evaluate $f'(1)=0$.
3) Evaluate $f''(1)=0$.
4) Evaluate $f'''(1)=0$.
5) Evaluate $f''''(1)=$:
$$
\frac{1}{2}f''''(1)=g_1'(1)g_2'(1)g_3(1)g_4(1)-g_1'(1)g_2(1)g_3'(1)g_4(1)-g_1(1)g_2'(1)g_3(1)g_4'(1)+g_1(1)g_2(1)g_3'(1)g_4'(1)
$$
With this in mind, and if you are certain that $f(x)$ is a fourth-degree polynomial, then we can say that
$$
f(x)=f(1)+(x-1)f'(1)+\frac{1}{2}(x-1)^2f''(1)+\frac{1}{3!}(x-1)^3f'''(1)+\frac{1}{4!}(x-1)^4 f''''(1)=
$$
$$
=\frac{1}{12}(x-1)^4 \bigg[g_1'(1)g_2'(1)g_3(1)g_4(1)-g_1'(1)g_2(1)g_3'(1)g_4(1)-g_1(1)g_2'(1)g_3(1)g_4'(1)+g_1(1)g_2(1)g_3'(1)g_4'(1)\bigg]
$$
I know this is not a very nice answer, but I believe its good enough for now, until any other one here post something better.
