If $f(x),g(x),h(x),\phi(x)$ are polynomials in $x$,$(\int_1^xf(x)h(x)dx)(\int_1^xg(x)\phi(x)dx)-(\int_1^xf(x)\phi(x)dx)(\int_1^xg(x)h(x)dx)$ If $f(x),g(x),h(x),\phi(x)$ are polynomials in $x$,$$(\int_1^xf(x)h(x)dx)(\int_1^xg(x)\phi(x)dx)-(\int_1^xf(x)\phi(x)dx)(\int_1^xg(x)h(x)dx)$$ is divisible by $(x-1)^t$.Then find the maximum value of $t$
I have no clue how to start attempting this.Answer given is $4$
 A: Let 
$$T(x)=\left(\int_1^xf(t)h(t)dt\right)\left(\int_1^xg(t)s(t)dt\right)-\left(\int_1^xf(t)s(t)dt\right)\left(\int_1^xg(t)h(t)dt\right).$$
Since $f,g,h,s$ are all polynomial functions so upon integration $T(x)$ will also be a polynomial in $x$. Furthermore, $T(1)=0$. This means $(x-1)$ is a factor of $T(x)$.
So to show that $(x-1)^2$ divides $T(x)$, we need to show that $T'(1)=0$. Likewise to show $(x-1)^4$ is a factor of $T(x)$, we need to show that $T^{'''}(1)=0$.
I will show it for the first derivative, after that you may fill in the details.
Using the fundamental theorem of calculus and product rule of derivatives, we get
$$T'(x)=\left(\int_1^xf(t)h(t)dt\right)\left[g(x)s(x)\right]+\left(\int_1^xg(t)s(t)dt\right)[f(x)h(x)]-\left(\int_1^xf(t)s(t)dt\right)\left[g(x)h(x)\right]-\left(\int_1^xg(t)h(t)dt\right)\left[f(x)s(x)\right].$$
Since each term involves the integral from $1$ to $x$. So $T'(1)=0$. This shows that $(x-1)^2$ is a factor of $T(x)$. 
Now continue with higher derivatives. 
