<s>[INST] Find $\int_0^bg(x,u)\,du$ given $\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$ I encounter the following equation, which holds for any $x\in[0,b]$:
$$\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$$
I would like to solve for $G(x,b)=\int_0^b g(x,u)du$, subject to the condition that $G(x,0)=0$. Intuitively, $G(x,b)$ is the column sum of an (infinite dimensional) symmetric matrix $g$ with entries given by the $g(x,y)$ function for $x,y\in[0,b]$, and the integral on the l.h.s. says that pre-multiplying the $x$-th column of $g$ by the vector $u^2+(b^2-1)u+(b-1)^2$ yields a known function of column sum in terms of $x$.
It seems that integration by parts won't really simplify the problem as the integral of $G$ then enters into the problem. I am not sure whether this question is solvable at all, and frankly, I do not even know where to start with. Maybe I have missed something very simple. Any hints or suggestions are highly appreciated.
 [/INST] We have \begin{align}\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)\,du&=\frac{x^2+b^2x+b^2-2b}{x-1}\tag1\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_x(x,u)\,du&=1-\frac{2b^2-2b+1}{(x-1)^2}\tag2\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_{xx}(x,u)\,du&=\frac{2(2b^2-2b+1)}{(x-1)^3}\end{align} so assume that $(x-1)^3g_{xx}(x,u)$ is constant. Then $g_{xx}(x,u)=t(b)/(x-1)^3$ where \begin{align}t(b)=\frac{2(2b^2-2b+1)}{\int_0^bu^2+(b^2-1)u+(b-1)^2\,du}=\frac{12(2b^2-2b+1)}{b(3b^3+8b^2-15b+6)}\end{align} so $$g_x(x,u)=h_1(u)-\frac{t(b)}{2(x-1)^2}\implies g(x,u)=h_2(u)+xh_1(u)+\frac{t(b)}{x-1}$$ where $h_1,h_2$ can be determined (with considerable freedom) using $(1)$ and $(2)$.
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